Equilibrium - Class 11 Chemistry - Chapter 6 - Notes, NCERT Solutions & Extra Questions
Renews every month. Cancel anytime
Extra Questions - Equilibrium | NCERT | Chemistry | Class 11
Strongest conjugate base is:
(A) $\mathrm{Cl}^-$ (B) $\mathrm{Br}^-$ (C) $\mathrm{F}^-$ (D) $\mathrm{I}^-$
The correct option is (C) $\mathrm{F}^-$.
The strength of a conjugate base is inversely related to the strength of its corresponding acid. Weaker acids have stronger conjugate bases.
Considering the acids formed by addition of an $\mathrm{H}$ atom to each base, we have:
$\mathrm{HF}$
$\mathrm{HCl}$
$\mathrm{HBr}$
$\mathrm{HI}$
Among these, the $\mathrm{F}^-$ ion is the smallest in size, causing its electron density to be more concentrated around its nucleus. This results in a short and stable $\mathrm{H}-\mathrm{F}$ bond. The stability and short bond length make the $\mathrm{H}-\mathrm{F}$ bond difficult to break, indicating a challenging $\mathrm{H}^+$ ion donation process. Therefore, $\mathrm{HF}$ represents the weakest acid among the list. Consequently, its conjugate base $\mathrm{F}^-$ is the strongest.
Alternate Explanation:
The smallest size and highest electronegativity of $\mathrm{F}$ help it gain protons effectively, further supporting the character of $\mathrm{F}^-$ as the strongest conjugate base: $$ \mathrm{H}^+ + \mathrm{F}^- \rightarrow \mathrm{HF} $$
$X Y_{2}$ dissociates as $X Y_{2}(g) \rightleftharpoons X Y(g) + Y(g)$ when the initial pressure of $X Y_{2}$ is $600 , \mathrm{mm}$ of $\mathrm{Hg}$. The total equilibrium pressure is $800 , \mathrm{mm}$ of $\mathrm{Hg}$. Find the value of the equilibrium constant for the reaction, assuming that the volume of the system remains unchanged.
A) $50 , \mathrm{mm} , \mathrm{Hg}$
B) $100 , \mathrm{mm} , \mathrm{Hg}$
C) $166.6 , \mathrm{mm} , \mathrm{Hg}$
D) $400 , \mathrm{mm} , \mathrm{Hg}$
Solution
The correct option is B) $100 , \mathrm{mm} , \mathrm{Hg}$.
The dissociation reaction can be represented as: $$ XY_{2}(g) \rightleftharpoons XY(g) + Y(g) $$
Given:
- Initial pressure of $XY_{2}$: $600 , \mathrm{mm, Hg}$
- Total equilibrium pressure: $800 , \mathrm{mm, Hg}$
Let $x , \mathrm{mm, Hg}$ be the amount of $XY_{2}$ that dissociates. This also creates $x , \mathrm{mm, Hg}$ each of $XY$ and $Y$. Therefore, the remaining pressure of $XY_{2}$ is $(600 - x) , \mathrm{mm, Hg}$.
The total pressure at equilibrium, considering the contributions from all species, is: $$ P_{total} = (600 - x) + x + x = 600 + x , \mathrm{mm, Hg} $$
Setting the total pressure equation to the equilibrium pressure: $$ 600 + x = 800 \quad \Rightarrow \quad x = 200 , \mathrm{mm, Hg} $$
At equilibrium, the pressures for the various components are:
- $P_{XY_{2}} = 600 - x = 400 , \mathrm{mm, Hg}$
- $P_{XY} = x = 200 , \mathrm{mm, Hg}$
- $P_{Y} = x = 200 , \mathrm{mm, Hg}$
The equilibrium constant $K_p$ is defined as: $$ K_p = \frac{P_{XY} \cdot P_{Y}}{P_{XY_{2}}} $$
Calculating $K_p$: $$ K_p = \frac{200 \times 200}{400} = 100 , \mathrm{mm, Hg} $$
Thus, the value of the equilibrium constant $K_p$ is $100 , \mathrm{mm, Hg}$.
For the system $3A + 2B \rightleftharpoons C$, the expression for equilibrium constant is
A $([3A][2B]) / [C]$
B $[C] / ([3A][2B])$
C $([A]^{3}[B]^{2}) / [C]$
D $[C] / ([A]^{3}[B]^{2})$
Solution
The correct answer is Option D: $$ [C] / ([A]^{3}[B]^{2}) $$
Explanation:
The equilibrium constant expression for the given reaction, $3A + 2B \rightleftharpoons C$, is constructed from the stoichiometry of the reactants and products. By the convention for writing equilibrium constant expressions in terms of concentration (denoted by $ K_c $), the expression is:
$$ K = \frac{[C]}{[A]^{3}[B]^{2}} $$
In this formula:
- The concentration of the product $C$ appears in the numerator.
- The concentrations of the reactants $A$ and $B$ appear in the denominator raised to powers equivalent to their coefficients in the balanced chemical equation.
Thus, the correct option that fits this description is Option D.
If, in the reaction $\mathrm{N}{2} \mathrm{O}{4} (\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}{2} (\mathrm{g})$, one mole of $\mathrm{N}{2} \mathrm{O}{4}$ is taken initially and $x$ is that part of $\mathrm{N}{2} \mathrm{O}_{4}$ which dissociates, then the number of moles at equilibrium will be:
A) 1
B) 3
C) $(1+x)$
D) $(1+x)^{2}$
The chemical reaction given is: $$ \mathrm{N}{2}\mathrm{O}{4} (\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2} (\mathrm{g}) $$
To find the number of moles at equilibrium, consider the dissociation process at the start ($t=0$) and at equilibrium ($t_{\text{eq}}$):
- Initially, there is 1 mole of $\mathrm{N}{2} \mathrm{O}{4}$ and 0 moles of $\mathrm{NO}_{2}$.
- At equilibrium, $x$ moles of $\mathrm{N}{2} \mathrm{O}{4}$ dissociate, leaving $(1-x)$ moles of $\mathrm{N}{2} \mathrm{O}{4}$.
- Each mole of $\mathrm{N}{2} \mathrm{O}{4}$ that dissociates produces 2 moles of $\mathrm{NO}{2}$. Therefore, $x$ moles of $\mathrm{N}{2} \mathrm{O}{4}$ produce $2x$ moles of $\mathrm{NO}{2}$.
To calculate the total number of moles at equilibrium: $$ (1-x) + 2x = 1 + x $$
Thus, the number of moles at equilibrium is $(1+x)$. Hence, the correct option is:
C) $(1+x)$
$\mathrm{I}{2}(\mathrm{aq}) + \mathrm{I}^{-}(\mathrm{aq}) \rightleftharpoons \mathrm{I}{3}^{-}(\mathrm{aq})$. We started with 1 mole of $\mathrm{I}{2}$ and 0.5 moles of $\mathrm{I}^{-}$ in a one-litre flask. After equilibrium is reached, an excess of $\mathrm{AgNO}{3}$ gave 0.25 moles of yellow precipitate. The equilibrium constant is:
A) 1.33 B) 2.66 C) 2 D) 3
The correct answer is A) 1.33.
The chemical reaction in question is: $$ \mathrm{I}_2 (\mathrm{aq}) + \mathrm{I}^- (\mathrm{aq}) \rightleftharpoons \mathrm{I}_3^- (\mathrm{aq}) $$
Initial concentrations are as follows: $\mathrm{I}_2 = 1 \text{ mol}$ and $\mathrm{I}^- = 0.5 \text{ mol}$.
To find the equilibrium constant $\mathrm{K}_c$, we analyze how the reaction progresses: $$ \mathrm{K}_c = \frac{[\mathrm{I}_3^-]}{[\mathrm{I}_2][\mathrm{I}^-]} $$
During the precipitation reaction with $\mathrm{AgNO_3}$: $$ \mathrm{I}^- + \mathrm{AgNO}_3 \rightarrow \underset{\text{yellow precipitate}}{\mathrm{AgI} \downarrow} + \mathrm{NO}_3^- $$ The formation of 0.25 moles of yellow precipitate (AgI) corresponds to 0.25 moles of $\mathrm{I}^-$ consumed.
This implies that there are: $$ 0.5 \text{ moles initial } \mathrm{I}^- - 0.25 \text{ moles consumed } = 0.25 \text{ moles remaining at equilibrium}. $$
Simultaneously, $\mathrm{I}_2$ also decreases by the stoichiometric equivalent of 0.25 moles: $$ 1 \text{ mole initial } \mathrm{I}_2 - 0.25 \text{ moles consumed } = 0.75 \text{ moles remaining}. $$
Therefore, $\mathrm{I}_3^-$ formed is: $$ 0.25 \text{ moles } (\text{same as the decrease of } \mathrm{I}_2 \text{ and } \mathrm{I}^-). $$
Substituting these equilibrium concentrations into the expression for $\mathrm{K}_c$: $$ \mathrm{K}_c = \frac{0.25}{0.75 \times 0.25} = \frac{0.25}{0.1875} = 1.33 $$
Thus, the equilibrium constant $\mathrm{K}_c$ for this reaction is 1.33.