Question

At T(K), \( K_c \) for the dissociation of \( PCl_5 \) is \( 2 \times 10^{-2} \) mol L\(^{-1}\). The number of moles of \( PCl_5 \) that must be taken in 1.0 L flask at the same temperature to get 0.2 mol of chlorine at equilibrium is

• A. \( 2.2 \)
• B. \( 1.1 \)
• C. \( 1.8 \)
• D. \( 4.4 \)

Hint:

Use the ICE table (Initial, Change, Equilibrium) method to solve equilibrium problems.
- The equilibrium constant expression is derived based on the balanced chemical equation.

Answer 1

The Correct Option is A

The dissociation reaction of \( PCl_5 \) is: \[ PCl_5 \rightleftharpoons PCl_3 + Cl_2 \] Let the initial moles of \( PCl_5 \) be \( x \). At equilibrium, if \( 0.2 \) mol of \( Cl_2 \) is formed, then the amount of \( PCl_5 \) dissociated will also be \( 0.2 \) mol, and the amount of \( PCl_3 \) formed will be \( 0.2 \) mol. Thus, at equilibrium: - \( PCl_5 \) left = \( x - 0.2 \) mol, - \( PCl_3 \) = \( 0.2 \) mol, - \( Cl_2 \) = \( 0.2 \) mol. The equilibrium constant expression is: \[ K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} \] Substituting the values: \[ 2 \times 10^{-2} = \frac{(0.2)(0.2)}{x - 0.2} \] \[ 2 \times 10^{-2} (x - 0.2) = 0.04 \] \[ x - 0.2 = \frac{0.04}{2 \times 10^{-2}} \] \[ x - 0.2 = 2 \] \[ x = 2.2 \] Thus, the correct answer is \(\boxed{2.2}\).