Question

The equilibrium $K_{c}$ for the reaction $\mathrm{SO}{2}(g) + \mathrm{NO}{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g) + \mathrm{NO}(g)$ is 16. If 1 mole of each of the four gases is taken in a $1 \mathrm{dm}^{3}$ vessel, the equilibrium concentration of NO would be:

A. $0.4 \mathrm{M}$
B. $0.6 \mathrm{M}$
C. $1 \mathrm{M}$

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Answer

To solve the given problem, let's go through the necessary steps methodically.

Given Data:

  • The reaction is: $$ \mathrm{SO}_2(g) + \mathrm{NO}_2(g) \rightleftharpoons \mathrm{SO}_3(g) + \mathrm{NO}(g) $$
  • Equilibrium constant ($K_c$) for the reaction is 16.
  • Initially, 1 mole of each of the four gases is taken in a 1 dm³ vessel (volume = 1 dm³ = 1 liter).

Initial Mole Calculation:

At time ( t = 0 ):

  • $\mathrm{SO}_2 = 1 \text{ mole}$
  • $\mathrm{NO}_2 = 1 \text{ mole}$
  • $\mathrm{SO}_3 = 1 \text{ mole}$
  • $\mathrm{NO} = 1 \text{ mole}$

Let's denote the change in moles at equilibrium as ( x ).

Change in Moles at Equilibrium:

  • $\mathrm{SO}_2 = 1 - x \text{ mole}$
  • $\mathrm{NO}_2 = 1 - x \text{ mole}$
  • $\mathrm{SO}_3 = 1 + x \text{ mole}$
  • $\mathrm{NO} = 1 + x \text{ mole}$

Concentration Calculations:

Since the volume of the vessel is 1 liter, the concentration of each gas will be:

  • $[\mathrm{SO}_2] = 1 - x$
  • $[\mathrm{NO}_2] = 1 - x$
  • $[\mathrm{SO}_3] = 1 + x$
  • $[\mathrm{NO}] = 1 + x$

Expression for Equilibrium Constant:

Using the equilibrium constant expression: $$ K_c = \frac{[\mathrm{SO}_3][\mathrm{NO}]}{[\mathrm{SO}_2][\mathrm{NO}_2]} $$ Substituting the equilibrium concentrations into the equation: $$ 16 = \frac{(1 + x)(1 + x)}{(1 - x)(1 - x)} $$ This simplifies to: $$ 16 = \left(\frac{1 + x}{1 - x}\right)^2 $$ Taking the square root of both sides: $$ 4 = \frac{1 + x}{1 - x} $$ Solving for ( x ): $$ 4(1 - x) = 1 + x \ 4 - 4x = 1 + x \ 4 - 1 = 4x + x \ 3 = 5x \ x = \frac{3}{5} = 0.6 $$

Equilibrium Concentration of NO:

The concentration of $\mathrm{NO}$ at equilibrium is $1 + x$: $$ [\mathrm{NO}] = 1 + 0.6 = 1.6 \text{ M} $$

Answer:

The equilibrium concentration of NO is 1.6 M, which corresponds to option D.


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