Thermodynamics - Class 11 Chemistry - Chapter 5 - Notes, NCERT Solutions & Extra Questions
Renews every month. Cancel anytime
Extra Questions - Thermodynamics | NCERT | Chemistry | Class 11
If 2 moles of an ideal monoatomic gas at temperature $T_{0}$ is mixed with 4 moles of another ideal monoatomic gas at temperature $2 \mathrm{~T}_{0}$, then the temperature of the mixture is
A) $\frac{5}{3} T_{0}$
B) $\frac{3}{2} T_{0}$
C) $\frac{4}{3} T_{0}$
D) $\frac{5}{4} T_{0}$
The correct answer is A) $\frac{5}{3} T_{0}$. This conclusion is found by considering the conservation of energy and the properties of ideal gases.
Firstly, since both gases are monoatomic, their degrees of freedom, denoted as $ f $, are similar. For an ideal monoatomic gas, the internal energy ($ U $) is given by: $$ U = \frac{n f R T}{2} $$ where $ n $ is the number of moles, $ R $ is the gas constant, and $ T $ is the temperature. The factor $ \frac{1}{2} $ accounts for the kinetic energy contribution in one degree of freedom (per mole per degree).
When the gases are mixed:
the first gas has 2 moles at temperature $ T_0 $,
the second gas has 4 moles at a temperature of $ 2 T_0 $.
The total internal energy of the mixture, by the conservation of energy, is the sum of the internal energies of the individual gases: $$ U_{\text{mix}} = U_1 + U_2 $$ We can substitute the values: $$ \begin{align*} U_{\text{mix}} &= \frac{2 f R T_0}{2} + \frac{4 f R (2 T_0)}{2} \ &= f R T_0 + 4 f R T_0 \ &= 5 f R T_0 \end{align*} $$ The internal energy of the mixture can also be written in terms of the unknown final temperature $ T_{\text{mix}} $ of the mixture: $$ U_{\text{mix}} = \frac{(n_1 + n_2) f R T_{\text{mix}}}{2} \ \Rightarrow \frac{(2 + 4) f R T_{\text{mix}}}{2} = 5 f R T_0 $$ Solving for $ T_{\text{mix}} $: $$ 6 f R T_{\text{mix}} = 10 f R T_0 \ T_{\text{mix}} = \frac{10 T_0}{6} = \frac{5}{3} T_0 $$ Thus, the temperature of the mixture is $\frac{5}{3} T_0$.
Two cylindrical rods of lengths $l_{1}$ and $l_{2}$, radii $r_{1}$ and $r_{2}$ have thermal conductivities $k_{1}$ and $k_{2}$ respectively. The ends of the rods are maintained at the same temperature difference. If $l_{1}=2l_{2}$ and $r_{1}=\frac{r_{2}}{2}$, the rates of heat flow in them will be the same if $\frac{k_{1}}{k_{2}}$ is
A. 1
B. 2
C. 4
D. 8
The correct answer is (D) 8.
The rate of heat flow for the two cylindrical rods can be represented by the formulas for rod A and rod B as: $$ \frac{Q_{1}}{t} = \frac{k_{1} \pi r_{1}^2 \Delta \theta}{l_{1}} $$ and $$ \frac{Q_{2}}{t} = \frac{k_{2} \pi r_{2}^2 \Delta \theta}{l_{2}}. $$ For the heat flow rates to be equal, we set $Q_1 = Q_2$, leading to: $$ \frac{k_{1} r_{1}^2}{l_{1}} = \frac{k_{2} r_{2}^2}{l_{2}}. $$ Given that $l_{1} = 2l_{2}$ and $r_{1} = \frac{r_{2}}{2}$, substituting these into our equation gives: $$ \frac{k_{1}}{k_{2}} = \frac{l_{1}}{l_{2}} \cdot \frac{r_{2}^2}{r_{1}^2} = 2 \cdot \left(\frac{r_{2}}{r_{1/2}}\right)^2 = 2 \cdot (2)^2 = 8. $$ Hence, $\frac{k_{1}}{k_{2}}$ must be 8 to ensure the heat flow rates are the same, making the correct option (D) 8.
On increasing the temperature, the kinetic energy of particles
A) Increases
B) Decreases
C) First increases then decreases.
D) Decreases rapidly.
The correct option is A) Increases
Temperature is directly related to the kinetic energy of particles. It is defined as a measure of the average kinetic energy of the particles in a substance. Therefore, when the temperature increases, the kinetic energy of the particles also increases.
For the reaction, $\mathrm{N}{2}(\mathrm{g})+3 \mathrm{H}{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{g}), \Delta \mathrm{H}$ is equal to: ($\triangle \mathrm{U} =$ change in internal energy)
A $\triangle U + 2 RT$
B $\triangle U - 2 RT$
C $\triangle U + RT$
D $\triangle U - RT$
The correct answer is $\mathbf{B} , \Delta U - 2RT$.
We apply the relation between enthalpy change ($\Delta H$) and internal energy change ($\Delta U$) in a chemical reaction, which is defined as: $$ \Delta H = \Delta U + \Delta n_{\text{g}} RT $$ where $\Delta n_{\text{g}}$ signifies the change in moles of gas.
For the given reaction: $$ \mathrm{N}2(\mathrm{g}) + 3\mathrm{H}2(\mathrm{g}) \rightleftharpoons 2\mathrm{NH}3(\mathrm{g}) $$ we calculate $\Delta n{\text{g}}$ as follows: $$ \Delta n{\text{g}} = (2) - (1 + 3) = 2 - 4 = -2 $$ Therefore, substituting into the enthalpy equation yields: $$ \Delta H = \Delta U + \Delta n{\text{g}} RT = \Delta U + (-2)RT = \Delta U - 2RT $$ Hence, the option that fits this calculation is Option B $\Delta U - 2RT$.
Assume the gas to be ideal, the work done on the gas in taking it from $A$ to $B$ is:
A) $200 \mathrm{R}$
B) $300 \mathrm{R}$
C) $400 \mathrm{R}$
D) $500 \mathrm{R}$
The correct answer is C) $400 \mathrm{R}$.
The work done on the gas as it transitions from point $A$ to point $B$ can be expressed by the equation: $$ W_{AB} = \Delta Q - \Delta U $$ where $W_{AB}$ is the work done on the gas, $\Delta Q$ is the heat added to the system, and $\Delta U$ is the change in internal energy. Utilizing the relationship between heat added at constant pressure and the change in internal energy, we get: $$ W_{AB} = nC_p \Delta T - nC_V \Delta T $$ This can be simplified, knowing that $C_p - C_V = R$ (by Mayer's relation): $$ W_{AB} = nR\Delta T $$ Given that there are $n = 2$ moles of the gas, and the change in temperature $\Delta T = (500 - 300)$ Kelvin is: $$ W_{AB} = 2 \cdot R \cdot 200 = 400R $$
Thus, the work done on the gas as it moves from point $A$ to point $B** is indeed $400 \mathrm{R}**.
The ionization enthalpies of lithium and sodium are 520 kJ/mole and 495 kJ/mole respectively. The energies required to convert all the atoms present in 7 mg of lithium vapours and 23 mg of sodium vapours to their respective gaseous cations are:
52 J, 49.5 J
520 J, 495 J
49.5 J, 52 J
495 J, 520 J
The correct option is B: $520 \text{ J}, 495 \text{ J}$
Calculation of moles:
For lithium ($\mathrm{Li}$): $$ \text{Number of moles of } \mathrm{Li} = \frac{7 , \text{mg}}{6.941 , \text{g/mol}} = \frac{7 \times 10^{-3}}{6.941} \approx 10^{-3} $$
For sodium ($\mathrm{Na}$): $$ \text{Number of moles of } \mathrm{Na} = \frac{23 , \text{mg}}{22.9898 , \text{g/mol}} = \frac{23 \times 10^{-3}}{22.9898} \approx 10^{-3} $$
Calculation of the energy required:
For lithium: $$ \text{Energy required for 1 mole of } \mathrm{Li} = 520 , \text{kJ/mol} $$ $$ \text{Energy required for } 10^{-3} \text{ moles of } \mathrm{Li} = 520 \times 10^{-3} = 520 \text{ J} $$
For sodium: $$ \text{Energy required for 1 mole of } \mathrm{Na} = 495 , \text{kJ/mol} $$ $$ \text{Energy required for } 10^{-3} \text{ moles of } \mathrm{Na} = 495 \times 10^{-3} = 495 \text{ J} $$
Therefore, the energies required to ionize 7 mg of lithium vapors and 23 mg of sodium vapors are 520 J and 495 J respectively.
4.48 L of an ideal gas at STP requires 12.0 calories to raise its temperature by 15°C at constant volume. What will be the value of Cp (in calorie).
To determine the value of ( C_p ) for an ideal gas given certain conditions, let's go through the step-by-step solution:
Given Information:
Volume of gas: 4.48 L
Temperature increase: 15°C
Heat provided: 12.0 calories
Conditions: STP (Standard Temperature and Pressure)
Determine the Number of Moles:
At STP, 1 mole of an ideal gas occupies ( 22.4 , \text{L} ).
We need to find the moles of gas in ( 4.48 , \text{L} ): $$ x = \frac{4.48 , \text{L}}{22.4 , \text{L/mol}} = 0.2 , \text{moles} $$
Calculate ( C_v ) (Specific Heat at Constant Volume):
Given, to raise the temperature of ( 0.2 ) moles of gas by ( 15°C ), it requires ( 12 ) calories.
To find the heat required to raise the temperature of 1 mole of gas by ( 1°C ): $$ C_v = \frac{12 , \text{calories}}{0.2 , \text{moles} \times 15 , °C} = \frac{12}{3} = 4 , \text{calories/mole °C} $$
Use the Relation Between ( C_p ) and ( C_v ):
For an ideal gas, ( C_p - C_v = R ), where ( R ) is the universal gas constant.
( R ) (in calories) = 2 calories/mole·°C.
Therefore, ( C_p ) is: $$ C_p = C_v + R = 4 , \text{calories/mole°C} + 2 , \text{calories/mole°C} = 6 , \text{calories/mole°C} $$
Answer: The value of ( C_p ) is 6 calories/mole·°C.
Choose the correct statement(s):
(A) Oxides of Ag and Hg can be reduced on heating
(B) $\text{CsL}_3 (s)$ is thermodynamically more stable than $\text{NaI}_3 (s)$
(C) Roasting is a cause of acid rain.
(D) Graphite is thermodynamically more stable than diamond
The correct options are:
(A) Oxides of $\mathrm{Ag}$ and $\mathrm{Hg}$ can be reduced on heating
(B) $\mathrm{CsI}_3(\mathrm{s})$ is thermodynamically more stable than $\mathrm{NaI}_3(\mathrm{s})$
(D) Graphite is thermodynamically more stable than diamond
Explanation:
(A) The oxides of metals like $\mathrm{Ag}$ and $\mathrm{Hg}$ can be reduced by heating. The reactions are as follows:
$$ \begin{array}{l} 2 \mathrm{HgO} \xrightarrow{\text{Heating}} 2 \mathrm{Hg} + \mathrm{O}_2 \ 2 \mathrm{Ag}_2 \mathrm{O} \xrightarrow{\text{Heating}} 4 \mathrm{Ag} + \mathrm{O}_2 \end{array} $$
(B) $\mathrm{CsI}_3(\mathrm{s})$ is thermodynamically more stable than $\mathrm{NaI}_3(\mathrm{s})$. This is primarily due to the fact that the cation and anion in $\mathrm{CsI}_3$ are of comparable sizes, whereas in $\mathrm{NaI}_3$, the large difference in sizes leads to lower stability.
(D) Graphite is thermodynamically more stable than diamond. Graphite contains one delocalized electron per carbon atom, which leads to greater attraction between carbon atoms, resulting in stronger bonds and more stability. Additionally, graphite exhibits greater van der Waals forces. In contrast, diamond does not contain delocalized electrons.
For the hydrogenation of $C_6H_6(l)$ to $C_6H_{12}(l)$, the hydrogenation enthalpy is -205 kJ/mol, and the resonance energy for $C_6H_6(l)$ is -152 kJ/mol. Therefore, the hydrogenation enthalpy for the following compound is?
Assume that the values of $\Delta H_{\text{vap}}$ for $C_6H_6(l)$, $C_6H_8(l)$, and $C_6H_{12}(l)$ are the same.
The correct answer is D.
For the hydrogenation of benzene, the theoretical hydrogenation enthalpy can be calculated using the following relationship:
Theoretical Hydrogenation Enthalpy = (Actual Hydrogenation Enthalpy) + (Resonance Energy)
Given:
Actual Hydrogenation Enthalpy of $C_6H_6(l)$ = -205 kJ/mol
Resonance Energy of $C_6H_6(l)$ = -152 kJ/mol
To find the hydrogenation enthalpy per double bond, we divide by 3, as benzene ($C_6H_6$) has three double bonds:
$$ \text{Hydrogenation Enthalpy} = \frac{-357}{3} = -119 \text{ kJ/mol} $$
Therefore, the theoretical enthalpy for the hydrogenation of benzene is -119 kJ/mol.
Final Answer: D
In a closed container, 100 mol of an ideal gas is taken. A movable, frictionless, weightless piston in this container moves in such a way that the pressure of the gas remains constant at 8.21 atm. Which graph of log $ V $ vs. log $ T $ (where $ V $ is the volume in litres and $ T $ is the temperature in Kelvin) is correct?
A.
B.
C.
D.
The correct answer is A.
Here's the detailed solution:
Given:
Number of moles of the ideal gas, $ n = 100 $ moles
The pressure of the gas, $ P = 8.21 $ atmospheres
Universal gas constant, $ R = 0.0821 , \text{L atm K}^{-1} \text{mol}^{-1} $
Using the ideal gas law: $$ PV = nRT $$
Substituting the given values: $$ V = \frac{nRT}{P} = \frac{100 \times 0.0821 \times T}{8.21} $$
Simplifying: $$ V = T $$
Taking the logarithm on both sides: $$ \log V = \log T $$
Thus, the graph of $\log V$ vs. $\log T$ will be a straight line with a slope of 1, which corresponds to option A.
Final Answer: A
In a mixture of gas $A$ ($C_{V, m}=\frac{3}{2} R$) and ($C_{V, m}=\frac{5}{2} R$), there are two moles of each present. The average molar heat capacity of the gaseous mixture at constant volume is:
A $R$
B $2R$
C $3R$
D $8R$
The correct answer is: B
To find the average molar heat capacity at constant volume ($C_{V,m}$) for the gas mixture, we use the formula:
$$ \text{औसत} , C_{V,m} = \frac{n_1 C_{V,m_1} + n_2 C_{V,m_2}}{n_1 + n_2} $$
Given:
For gas $A$: $C_{V,m_1} = \frac{3}{2}R$, $n_1 = 2$ moles
For gas $B$: $C_{V,m_2} = \frac{5}{2}R$, $n_2 = 2$ moles
Substitute these values into the formula:
$$ \text{औसत} , C_{V,m} = \frac{2 \times \frac{3}{2} R + 2 \times \frac{5}{2} R}{2 + 2} $$
Simplify the expression:
$$ = \frac{2 \left(\frac{3}{2}R\right) + 2 \left(\frac{5}{2}R\right)}{4} $$
$$ = \frac{3R + 5R}{4} $$
$$ = \frac{8R}{4} $$
$$ = 2R $$
Therefore, the average molar heat capacity at constant volume for the gas mixture is $2R$, which corresponds to option B.
In the context of entropy, the correct statement is:
At absolute zero temperature, the entropy of a perfectly crystalline solid is zero.
At $0^{\circ} \mathrm{C}$, the entropy of a perfectly crystalline substance is zero.
At absolute zero temperature, the entropy of a perfectly crystalline solid is zero.
At absolute zero temperature, the entropy of all crystalline substances is zero.
Correct Answer: A
According to the third law of thermodynamics, the entropy of a perfect crystalline solid is zero at absolute zero temperature.
Therefore, the right statement is:
At absolute zero temperature, the entropy of a perfect crystalline solid is zero.
Final Answer: A
Consider the following reaction:
$$ \mathrm{C}_6\mathrm{H}_6(\ell) + \frac{15}{2}\mathrm{O}_2(\text{g}) \rightarrow 6\mathrm{CO}_2(\text{g}) + 3\mathrm{H}_2\mathrm{O}(\text{g}) $$
For the above reaction, the signs of $\Delta H, \Delta S$, and $\Delta G$ are respectively:
A $ \quad +, -, + $
B -, - +
C -, - ++
D, ++ -
Let's analyze the given reaction:
$$ \mathrm{C}_6\mathrm{H}_6(\ell) + \frac{15}{2}\mathrm{O}_2(\text{g}) \rightarrow 6\mathrm{CO}_2(\text{g}) + 3\mathrm{H}_2\mathrm{O}(\text{g}) $$
Key points to consider for $\Delta H$, $\Delta S$, and $\Delta G$:
Enthalpy Change ($\Delta H$):
Combustion reactions are typically exothermic.
Hence, $\Delta H$ for this reaction is negative ($-ve$).
Entropy Change ($\Delta S$):
Calculate $\Delta n_g$: the change in the number of moles of gases.
Reactants: $\frac{15}{2}\mathrm{O}_2$ = 7.5 moles (only considering gaseous states)
Products: $6\mathrm{CO}_2 + 3\mathrm{H}_2\mathrm{O}$ = 9 moles
$\Delta n_g = 9 - 7.5 = +1.5$ (positive change)
Therefore, $\Delta S$ is positive ($+ve$).
Gibbs Free Energy Change ($\Delta G$):
$\Delta G = \Delta H - T \Delta S$
With $\Delta H$ being negative and $\Delta S$ being positive, $\Delta G$ will also be negative ($-ve$).
Thus, for this reaction, the signs of $\Delta H$, $\Delta S$, and $\Delta G$ are negative, positive, and negative respectively.
The correct answer is B.
Statement of Hess's Law:
The standard enthalpy of a complete reaction is equal to the sum of the enthalpies of the various reactions participating in it.
The enthalpy of formation of a compound is equal to its decomposition enthalpy into its elements, but the signs are opposite.
At constant temperature, the pressure of a gas is proportional to the mass of the gas dissolved per unit volume of a solvent.
The pressure of the gas is directly proportional to its concentration in the solution and is inversely proportional to the volume of the solution in equilibrium.
Final Answer: A
According to Hess's Law, the standard enthalpy of a complete reaction is equal to the sum of the enthalpies of the individual reactions that compose it.
Explanation: According to Hess's Law, the standard enthalpy of a complete reaction is equal to the sum of the enthalpies of the individual reactions that compose it. This is a practical method in thermodynamics for determining the enthalpy change of a complex reaction by breaking it into simpler steps whose enthalpies are known.
At $298 \text{ K}$, the standard enthalpy change for gaseous $\text{H}_2\text{O}$ is -241.82 kJ/mol. The molar heat capacities (at constant pressure) of some substances are given below:
Assuming that the heat capacities do not depend on temperature, the value of $\Delta H^{\circ}$ at $373 \text{ K}$ is: A -242.6 kJ/mol B -485.2 kJ/mol C -121.3 kJ/mol D -286.4 kJ/mol
The correct answer is A.
We are given the standard enthalpy change for gaseous $\text{H}_2\text{O}$ at $298 \text{ K}$ is $$ -241.82 , \text{kJ/mol} $$.
The reaction is:
$$ \mathrm{H}{2}(\mathrm{g}) + \frac{1}{2} \mathrm{O}{2}(\mathrm{g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) $$
According to Kirchhoff's Law, the change in enthalpy at different temperatures can be expressed as:
$$ \Delta_{r} H_{T_{2}}^{\circ} - \Delta_{r} H_{T_{1}}^{\circ} = \Delta_{r} C_{P}^{\circ} \left[T_{2} - T_{1}\right] $$
Here, we have:
$\Delta_{r} H_{T_{1}}^{\circ} = -241.82 , \text{kJ/mol}$ (at $T_1 = 298 \text{ K}$)
$\Delta_{r} H_{T_{2}}^{\circ}$ (at $T_2 = 373 \text{ K}$)
$\Delta_{r} C_P^{\circ} = -10.945 , \text{J/K}$
Substituting into Kirchhoff's Law:
$$ \Delta_{r} H_{T_{2}}^{\circ} - (-241.82) = (-10.945) \left[373 - 298\right] $$
Calculate the temperature difference:
$$ 373 - 298 = 75 , \text{K} $$
Therefore,
$$ (-10.945) \times 75 = -820.875 , \text{J} $$
Since $1 , \text{kJ} = 1000 , \text{J}$, we convert to kJ:
$$ \Delta_{r} H_{T_{2}}^{\circ} - (-241.82) = \frac{-820.875 , \text{J}}{1000} = -0.820875 , \text{kJ} $$
So,
$$ \Delta_{r} H_{T_{2}}^{\circ} = -241.82 - 0.820875 = -242.64 , \text{kJ/mol} $$
Hence, the enthalpy change at $373 \text{ K}$ is approximately: $$ \boxed{-242.6 , \text{kJ/mol}} $$
Final Answer: A
The enthalpy change for the reaction $\mathrm{C}_2\mathrm{H}_6(\mathrm{g}) \rightarrow 2\mathrm{C}(\mathrm{g}) + 6\mathrm{H}(\mathrm{g})$ is $x$ kilojoules. The bond energy of the $\mathrm{C}-\mathrm{H}$ bond is:
A. $\frac{x}{2}$
B. $\frac{x}{3}$
C. $\frac{x}{6}$
D. Information is insufficient.
To determine the bond energy of $\mathrm{C}-\mathrm{H}$ bonds from the given reaction:
$$ \mathrm{C}_2\mathrm{H}_6(\mathrm{g}) \rightarrow 2\mathrm{C}(\mathrm{g}) + 6\mathrm{H}(\mathrm{g}) $$
we are provided that the enthalpy change for the reaction is $ x $ kJ. However, the energy associated with the carbon-carbon bond ($ \mathrm{C}-\mathrm{C} $ bond enthalpy) is not given.
Without knowing the enthalpy of the $ \mathrm{C}-\mathrm{C} $ bond, we cannot isolate and determine the bond energy of the $ \mathrm{C}-\mathrm{H} $ bonds. This makes it impossible to calculate the precise $ \mathrm{C}-\mathrm{H} $ bond energy using just the given data.
Hence, the correct answer is:
D. Information is insufficient.
At $300$ K, 1 mole of ideal gas has expanded isothermally from 10 atm to 1 atm. Given that gas constant $=2$ cal/mol K.
The work done is:
A. 938.8 calories
B. 1138.8 calories
C. 1381.8 calories
D. 15818 calories
To determine the work done during the isothermal expansion of an ideal gas, we can use the following formula:
$$ w = -2.303 nRT \log\left(\frac{P_1}{P_2}\right) $$
Given:
The temperature $T = 300 \text{ K}$
The number of moles $n = 1$
The gas constant $R = 2 \text{ cal/mol K}$
The initial pressure $P_1 = 10 \text{ atm}$
The final pressure $P_2 = 1 \text{ atm}$
We can substitute these values into the formula:
$$ w = -2.303 \times 1 \times 2 \times 300 \log\left(\frac{10}{1}\right) $$
This simplifies to:
$$ w = -2.303 \times 600 \log(10) $$
Given that $\log(10) = 1$, we get:
$$ w = -2.303 \times 600 $$
Calculating the value:
$$ w = -1381.8 \text{ cal} $$
Therefore, the correct option is:
C. 1381.8 cal
When a sample of gas was compressed from $500 ~\text{cm}^3$ to $300 \text{cm}^3$ by an average pressure of 0.6 atmospheres, 10 calories of heat was released in the process. The change in internal energy is:
A. $-2.16 ~\text{joule}$
B. $12.156 ~\text{joule}$
C. $2.156~\text{joule}$
D. $101.3~\text{joule}$
The correct answer is:C
To find the change in internal energy, we use the following steps:
Work Done Calculation:
The work done on the gas sample is given by: $$ w = P \Delta V $$
Given values: $\Delta V = 500 , \text{cc} - 300 , \text{cc} = 200 , \text{cc} = 0.2 , \text{liters}$, $P = 0.6 , \text{atm}$
Therefore: $$ w = 0.6 , \text{atm} \times 0.2 , \text{liters} = 0.12 , \text{liter-atm} $$
Converting Work Done to Joules:
Since $1 , \text{liter-atm}$ is equivalent to $101.3 , \text{Joules}$:
$$ w = 0.12 , \text{liter-atm} \times 101.3 , \text{Joules/liter-atm} = 12.156 , \text{Joules} $$Calculating Change in Internal Energy:
The internal energy change $( \Delta U )$ can be found considering the heat released $( -10 , \text{J} )$ and the work done on the gas $( 12.156 , \text{J} )$:
$$ \Delta U = -10 , \text{J} + 12.156 , \text{J} = 2.156 , \text{J} $$
Thus, the change in internal energy is 2.156 Joules.
Final Answer: C
Which reaction will have the highest value of $\Delta S$?
A $\mathrm{Ca}_{(s)} + \frac{1}{2} {\mathrm{O}_2}_{(g)} \rightarrow \mathrm{CaO}_{(s)}$
B ${\mathrm{CaCO}_3}_{(s)} \rightarrow \mathrm{CaO}_{(s)} + {\mathrm{CO}_2}_{(g)}$
C $\mathrm{C}_{(s)} + {\mathrm{O}_2}_{(g)} \rightarrow {\mathrm{CO}_2}_{(g)}$
D ${\mathrm{N}_2}_{(g)} + {\mathrm{O}_2}_{(g)} \rightarrow 2 \mathrm{NO}_{(g)}$
The correct answer is:
B
In the reaction where a solid substance converts into a gaseous substance, the value of $\Delta S$ is maximum.
Note that in option B, $\mathrm{CaCO}_{3(s)}$ converts from solid to $\mathrm{CO}_{2(g)}$ gas, which results in an increase in entropy.
The following data were obtained for the melting of compound $AB$: $\Delta H = 9.2 , \text{kJ mol}^{-1}$, $\Delta S = 0.008 , \text{kJ} , K^{-1} , \text{mol}^{-1}$. Its melting point is:
Options:
A. 736 K
B. 1050 K
C. 1150 K
D. 1150°C
To find the melting point of compound $AB$, we use the relationship:
$$ T = \frac{\Delta H_{\text{fusion}}}{\Delta S_{\text{fusion}}} $$
Given:
$\Delta H_{\text{fusion}} = 9.2$ kJ/mol
$\Delta S_{\text{fusion}} = 0.008$ kJ/(K mol)
Substitute the given values into the equation:
$$ T = \frac{9.2}{0.008} = 1150 \text{ K} $$
Thus, the melting point of compound $AB$ is 1150 K.
Final Answer: C
The reaction: $\mathrm{H}_2 + \mathrm{Cl}_2 \rightarrow 2 \mathrm{HCl}$, has $\Delta \mathrm{H}=194$ kilojoules. The enthalpy of formation of $\mathrm{HCl}$ is:
A +97 kilojoules
B +194 kilojoules
C -194 kilojoules
D -97 kilojoules
For the reaction: $\mathrm{H}_{2} + \mathrm{Cl}_{2} \rightarrow 2 \mathrm{HCl}$, the given $\Delta \mathrm{H}$ is 194 kJ.
To find the enthalpy of formation for $\mathrm{HCl}$, we use the relationship given:
$$ \mathrm{H}_{2} + \mathrm{Cl}_{2} \rightarrow 2 \mathrm{HCl} $$
Here, $\Delta \mathrm{H}$ for the formation of 2 moles of $\mathrm{HCl}$ is 194 kJ.
Therefore, the enthalpy of formation for 1 mole of $\mathrm{HCl}$ is:
$$ \text{Enthalpy of formation} = \frac{194 \text{ kJ}}{2} = 97 \text{ kJ} $$
Thus, the enthalpy of formation of $\mathrm{HCl}$ is +97 kJ.
Final Answer: A
For the neutralization reaction of acetic acid ($ \text{CH}_3 \text{COOH} $) and NaOH, the enthalpy of neutralization is $ -50.6 $ kJ/mol. For a strong acid and a strong base, the enthalpy of neutralization is $ -55.9 $ kJ/mol. The value of $ \Delta H $ for the ionization of $ \text{CH}_3 \text{COOH} $ is:
A +5.3 kJ/mol
B +6.2 kJ/mol
C +8.2 kJ/mol
D +9.3 kJ/mol
The correct answer is A.
To calculate the ionization enthalpy ($\Delta H$) for acetic acid $(\text{CH}_3\text{COOH})$, we start with the given data:
Neutralization enthalpy of acetic acid with NaOH: $$\Delta H_{\text{neutralization}} = -50.6 \text{ kJ/mol}$$
Neutralization enthalpy of a strong acid with a strong base: $$\Delta H_{\text{neutralization,strong}} = -55.9 \text{kJ/mol}$$
The ionization enthalpy of acetic acid ($\Delta H_{\text{ionization}}$) can be found using the following relationship:
$$-50.6 = \Delta H_{\text{ionization}} + (-55.9)$$
Rearrange to solve for $\Delta H_{\text{ionization}}$:
$$\Delta H_{\text{ionization}} = -50.6 + 55.9$$ $$\Delta H_{\text{ionization}} = +5.3 \text{kJ/mol}$$
Thus, the ionization enthalpy for CH$_3$COOH is +5.3 kJ/mol.
Final Answer: A
Which of the following does the entropy decrease in?
Crystallization of solute from solution
Rusting of iron
Melting of ice
Evaporation of camphor
The change in entropy in various processes occurs as follows:
Crystallization of solute from solution: In this process, an orderly structure (crystal) forms, which decreases entropy.
Rusting of iron: In this process, the randomness and disorder of iron molecules increase, which increases entropy.
Melting of ice: When ice melts and turns into water, the randomness of its molecules increases, which increases entropy.
Evaporation of camphor: In this process, camphor particles become more free and convert into gas, which increases entropy.
Therefore, crystallization of solute from solution is the only process in which entropy decreases.
Hence, the correct answer is:
Crystallization of solute from solution
Final Answer: A
The enthalpy change $ \Delta H = -189 $ kilojoules for the polymorphic transformation "C (diamond) → C (graphite)". If 6 grams of diamond and 6 grams of graphite are burnt separately to obtain $ \text{CO}_2 $, then the free energy in the first reaction:
A. will be 1.89 kilojoules less than the second equation. B. will be 1.89 kilojoules more than the second equation. C. will be 11.34 kilojoules less than the second equation. D. will be 0.945 kilojoules more than the second equation.
To determine the difference in heat released when burning 6 grams of diamond vs. 6 grams of graphite, we need to use the given enthalpy change for the conversion from diamond to graphite.
The enthalpy change for the process $$ C (\text{diamond}) \rightarrow C (\text{graphite}) $$ is given as $$ \Delta H = -189 , \text{kJ} $$.
First, let's calculate the number of moles of diamond and graphite:
The molar mass of carbon (C) is $ 12 , \text{g/mol} $.
For 6 grams of diamond: $$ \text{Moles of diamond} = \frac{ 6 , \text{g} }{ 12 , \text{g/mol} } = 0.5 , \text{mol} $$.
Similarly, for 6 grams of graphite: $$ \text{Moles of graphite} = \frac{ 6 , \text{g} }{ 12 , \text{g/mol} } = 0.5 , \text{mol} $$.
According to the enthalpy change, for the conversion of 1 mole of diamond to graphite, the energy change is $$ 189 , \text{kJ} $$. Thus, for 0.5 moles, it would be: $$ 0.5 \times 189 = 94.5 , \text{kJ} $$.
Because the enthalpy change is negative, this means that diamond is less stable than graphite by 94.5 kJ for 0.5 moles.
Therefore:
When burning 6 grams (0.5 moles) of diamond, it will release 94.5 kJ more energy than burning 6 grams (0.5 moles) of graphite.
So, in conclusion:
The energy released during the burning of diamond will be $$ 0.945 , \text{kJ} $$ more than that released during the burning of graphite.
Final Answer: D
In the adiabatic expansion of an ideal gas:
A. $ w = -\Delta U $
B. $ w = \Delta U $
C. $ \Delta U = 0 $
D. $ w = 0 $
The correct answer is: B
In thermodynamics, the relationship between heat ($q$), internal energy change ($\Delta U$), and work done ($w$) is given by:
$$ q = \Delta U - w $$
For an adiabatic process, the heat exchanged is zero ($q = 0$). Substituting this into the equation, we get:
$$ 0 = \Delta U - w $$
Therefore, the change in internal energy ($\Delta U$) is equal to the work done ($w$):
$$ \Delta U = w $$
Final Answer: B
When 0.5 grams of $ S $ is burned in $ SO_2 $, 4.6 kilojoules of heat is released. The enthalpy of formation for $ SO_2 $ is:
A. $+147.2 ~\text{kilojoule}$
B. $-147.2 ~\text{kilojoule}$
C. $+294.4 ~\text{kilojoule}$
D. $-294.4 ~\text{kilojoule}$
Given:
When 0.5 grams of S are burned in SO₂, 4.6 kJ of heat is released.
We need to calculate the standard enthalpy of formation for $ \text{SO}_2 $.
The reaction is: $$ S + O_2 \rightarrow SO_2, \Delta H_f = -4.6 \text{ kJ} $$
Since 0.5 grams of sulfur yields 1 gram of SO₂:
To find out how much heat is released per mole of sulfur, we use: $$ 32 \text{ grams of sulfur } = \left(\frac{32 \times 1}{0.5}\right) = 64 \text{ grams of } SO_2 $$
Calculate the enthalpy change: $$ \Delta H_f = -4.6 \text{ kJ} \times 64 = -294.4 \text{ kJ} $$
Thus, the standard enthalpy of formation for $ \text{SO}_2 $ is -294.4 kJ.
Final Answer: C
The standard entropies of $\mathrm{CO_2 (g)}$, $\mathrm{C (s)}$, and $\mathrm{O_2 (g)}$ are 213.5, 5.690, and 205 joules $\mathrm{K}^{-1}$ respectively. The standard Gibbs entropy of $\mathrm{CO_2 (g)}$ is:
A. 1.86 joules $\mathrm{K}^{-1}$
B. 1.96 joules $\mathrm{K}^{-1}$
C. 2.81 joules $\mathrm{K}^{-1}$
D. 2.86 joules $\mathrm{K}^{-1}$
To determine the standard entropy of formation for $\mathrm{CO}_{2(\mathrm{g})}$, consider the reaction:
$$ \mathrm{C_{(s)}} + \mathrm{O_{2(g)}} \rightarrow \mathrm{CO_2(g)} $$
Given the standard entropy values:
$\mathrm{CO}_{2(\mathrm{g})}$: 213.5 J K-1
$\mathrm{C}_{(\mathrm{s})}$: 5.690 J K-1
$\mathrm{O}_{2(\mathrm{g})}$: 205 J K-1
Using the formula for the change in entropy, $\Delta S^{\circ}$, we calculate:
$$ \Delta S^{\circ} = S^{\circ}(\mathrm{CO_2(g)}) - S^{\circ}(\mathrm{C_{(s)}}) - S^{\circ}(\mathrm{O_2(g)}) $$
Substituting the given values:
$$ \Delta S^{\circ} = 213.5 - 5.690 - 205 = 2.81 , \text{J K}^{-1} $$
Thus, the standard entropy of formation, $\Delta S^{\circ}$, for $\mathrm{CO}_{2(\mathrm{g})}$ is 2.81 J K-1.
Final Answer: C
Which reaction has $\Delta U = \Delta H$?
A. $\mathrm{N}_2\mathrm{O}_4{(g)} \Leftrightarrow 2\mathrm{NO}_2{(g)}$
B. $2\mathrm{SO}_2{(g)} + \mathrm{O}_2{(g)} \Leftrightarrow 2\mathrm{SO}_3{(g)}$
C. $\mathrm{H}_2{(g)} + \mathrm{Cl}_2{(g)} \Leftrightarrow 2\mathrm{HCl}{(g)}$
D. $\mathrm{H}_2{(g)} + \frac{1}{2}\mathrm{O}_2{(g)} \Leftrightarrow \mathrm{H}_2\mathrm{O}{(l)}$
The correct answer is: C
For the reaction: $\mathrm{H}_{2(g)} + \mathrm{Cl}_{2(g)} \Leftrightarrow 2\mathrm{HCl}_{(g)}$
Change in the number of moles of gas, $\Delta n = 0$
Hence, $\Delta U = \Delta H$.
Final Answer: C
What will be the value of ∆n for the combustion of 1 mole of benzene if both reactants and products are gaseous at 298 K?
A. 0
B. $\frac{3}{2}$
C. $-\frac{3}{2}$
D. $\frac{1}{2}$
The correct answer is:D
Let's go through the balanced chemical equation for the combustion of benzene:
$$ \mathrm{C}_6 \mathrm{H}_6 (g) + \frac{15}{2} \mathrm{O}_2 (g) \rightarrow 6 \mathrm{CO}_2 (g) + 3 \mathrm{H}_2 \mathrm{O} (g) $$
To find the change in the number of moles of gas ($ \Delta n $), we use the formula:
$$ \Delta n = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} $$
Substituting the values:
$$ \Delta n = (6 + 3) - \left(1 + \frac{15}{2}\right) $$
Simplify the expression:
$$ \Delta n = 9 - \left(1 + 7.5\right) $$
$$ \Delta n = 9 - 8.5 $$
$$ \Delta n = \frac{1}{2} $$
Therefore, the correct value of $\Delta n$ is $\frac{1}{2}$. Hence, the final answer is D.
In order to calculate the work done in joules for the reversible isothermal expansion of an ideal gas, the volume should be taken in which unit?
A. Only cubic meters $ \text{m}^3 $
B. Only liters $ \text{L} $
C. Only cubic centimeters $ \text{cm}^3 $
D. None of the above
The correct answer is:
D
The work done in a reversible isothermal expansion of an ideal gas can be calculated using the formula:
$$ W_{\text{rev}} = 2.303 , R , T , \log \left( \frac{V_2}{V_1} \right) $$
Since the formula uses the ratio $\frac{V_2}{V_1}$, the volume can be in any unit because the ratio itself is unitless.
(Note: The gas constant $R$ should be in Joules per Kelvin per mole ($\text{J} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}$)).
Thus, the correct option for the unit of volume in this context is:
D: None of the above
Below are two thermochemical equations:
$$ M + \frac{1}{2} O_{2} \rightarrow MO + 351.4 \text{ kilojoules} $$
$$ X + \frac{1}{2} O_{2} \rightarrow XO + 90.8 \text{ kilojoules} $$
Based on these, the heat of reaction for the process
$$ M + XO \Leftrightarrow MO + X $$
is:
A. 422.2 kilojoules
B. 268.7 kilojoules
C. -442.2 kilojoules
D. 260.6 kilojoules
To determine the reaction enthalpy for the process $M + XO \Leftrightarrow MO + X$, let us consider the given thermochemical equations:
$M + \frac{1}{2} O_{2} \rightarrow MO + 351.4$ kJ
$X + \frac{1}{2} O_{2} \rightarrow XO + 90.8$ kJ
We need to find $\Delta H$ for the reaction: $$ M + XO \Leftrightarrow MO + X $$
By subtracting equation (2) from equation (1), we get:
$$ \Delta H = (351.4 , \text{kJ}) - (90.8 , \text{kJ}) = 260.6 , \text{kJ} $$
Thus, the reaction enthalpy $\Delta H$ is 260.6 kJ.
Final Answer: D
The value of $\Delta S$ for the transformation $C_{\text{(graphite)}} \rightarrow C_{\text{(diamond)}}$ is:
A. Zero
B. Positive
C. Negative
D. Unknown
The correct answer is: C
The transformation of graphite to diamond is an endothermic reaction. This means that the entropy ($ \Delta S $) of diamond is lower than that of graphite. Therefore, for the process $$\text{graphite} \rightarrow \text{diamond}$$, the value of $\Delta S$ is negative. Hence, the correct answer is C.
For the reaction $\mathrm{N_2 + 3 H_2 \Leftrightarrow 2 NH_3}$, what is $\Delta H$?
A. $\Delta U - R T$
B. $\Delta U - 2 R T$
C. $\Delta U + R T$
D. $\Delta U + 2 R T$
The correct answer is B.
Given the reaction: $$ \mathrm{N}_{2} + 3 \mathrm{H}_{2} \Leftrightarrow 2 \mathrm{NH}_{3} $$
The relationship between the change in enthalpy ($ \Delta H $) and the change in internal energy ($ \Delta U $) is given by: $$ \Delta H = \Delta U + \Delta n R T $$
Here, $ \Delta n $ represents the change in the number of moles of gas. For this reaction:
Initial moles of reactants ($ \mathrm{N}_{2} $ and $ \mathrm{H}_{2} $): $1 + 3 = 4$
Final moles of products ($ \mathrm{NH}_{3} $): $2$
Thus, the change in moles of gas is: $$ \Delta n = 2 - 4 = -2 $$
Substituting this into the equation, we get: $$ \Delta H = \Delta U - 2 R T $$
Thus, the correct option is B.
For the reaction $2 \mathrm{NH}_{3(g)} \rightarrow N_{2(g)} + 3 H_{2(g)}$ at $27^{\circ} \mathrm{C}$, the value of $\Delta H - \Delta U$ is:
A $8.314 \times 273 \times (-2)$
B $8.314 \times 300 \times (-2)$
C $8.314 \times 27 \times (-2)$
D $8.314 \times 300 \times (2)$
To determine the value of $ \Delta H - \Delta U $ for the reaction $ 2 \mathrm{NH}_{3(g)} \rightarrow N_{2(g)} + 3 H_{2(g)} $ at $27^{\circ} \mathrm{C}$, we'll follow these steps:
Key Formula
$$ \Delta H - \Delta U = \Delta n \cdot R \cdot T $$
Steps:
Calculate the change in moles of gas, $ \Delta n $:
$$ \Delta n = \sum \text{(moles of gas products)} - \sum \text{(moles of gas reactants)} $$
For this reaction: $ \Delta n = (1 + 3) - 2 = 4 - 2 = 2 $Use the ideal gas constant, ( R ): $ R = 8.31 , \text{J/mol K} $
Temperature, ( T ):$ T = 27^\circ , \text{C} = 300 , \text{K} $
Substitute the values into the formula: $ \Delta H - \Delta U = 2 \times 8.31 \times 300 $
Calculation:
$$ 2 \times 8.31 \times 300 = 4986 , \text{J/mol} $$
Thus, the correct answer is: $ \boxed{D , (8.314 \times 300 \times (2))} $
The unit of entropy is:
A: Joule per mole $^{-1}$
B: Joule $\mathrm{K}$ per mole $^{-1}$
C: Joule per mole $^{-1} , \mathrm{K}^{-1}$
D: Joule $^{-1} , \mathrm{K}^{-1}$ per mole $^{-1}$
The correct answer is C.
The entropy change, $\Delta S$, is given by the equation: $$ \Delta S = \frac{q_{\mathrm{rev}}}{T} $$ where $q_{\mathrm{rev}}$ is the reversible heat exchange and $T$ is the temperature in Kelvin.
Therefore, the unit of entropy, $S$, is: $$ \text{Joules per Kelvin} , \left( \text{J} , \text{K}^{-1} \right) $$ Hence, the unit of entropy is joules per Kelvin per mole $joules$ $\text{K}^{-1}$ , $\text{mole}^{-1}$).
Final Answer: C
Two different adiabatic paths for the same gas intersect two isothermals at $T_{1}$ and $T_{2}$ as shown in the $P-V$ diagram. Then $\frac{V_{a}}{V_{d}}$ is equal to
A $\frac{V_{c}}{V_{b}}$
B $V_{c} V_{b}$
C $\frac{V_{b}}{V_{c}}$
D $\frac{V_{c} V_{b}}{V_{c}+V_{b}}$
The correct option is C: $\frac{V_{b}}{V_{c}}$.
Explanation:
For the curve $bc$ (an adiabatic process): [ T_{1}V_{b}^{\gamma-1} = T_{2}V_{c}^{\gamma-1} ]
For the curve $ad$ (another adiabatic process): [ T_{1}V_{a}^{\gamma-1} = T_{2}V_{d}^{\gamma-1} ]
Now, dividing the first equation by the second, we get: [ \frac{T_{1}V_{b}^{\gamma-1}}{T_{1}V_{a}^{\gamma-1}} = \frac{T_{2}V_{c}^{\gamma-1}}{T_{2}V_{d}^{\gamma-1}} ]
This simplifies to: [ \left( \frac{V_{b}}{V_{a}} \right)^{\gamma-1} = \left( \frac{V_{c}}{V_{d}} \right)^{\gamma-1} ]
Since the $\gamma-1$ exponentiation on both sides is equal, we can further simplify it to: [ \frac{V_{b}}{V_{a}} = \frac{V_{c}}{V_{d}} ]
Rearranging gives: [ \frac{V_{a}}{V_{d}} = \frac{V_{b}}{V_{c}} ]
Thus, $\frac{V_{a}}{V_{d}}$ is equal to $\frac{V_{b}}{V_{c}}$.