Question

One mole of \( PCl_5(g) \) was heated in a 1L closed flask at 500 K. At equilibrium, 0.1 mole of \( Cl_2(g) \) was formed. What is its \( K_p \) (in atm)? (Given R = 0.082 L atm mol\(^{-1} \) K\(^{-1} \))

• A. \( 2.7 \times 10^{-4} \)
• B. \( 0.455 \)
• C. \( 0.0111 \)
• D. \( 90.0 \)

Hint:

Set up an ICE (Initial, Change, Equilibrium) table to find the equilibrium concentrations of the reactants and products. Calculate \( K_c \) using these concentrations. Then use the relation \( K_p = K_c (RT)^{\Delta n_g} \) to find \( K_p \), where \( \Delta n_g \) is the change in the number of moles of gaseous species.

Answer 1

The Correct Option is B

The equilibrium reaction is: $$ PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) $$ Initial moles of \( PCl_5 \) = 1 mol Initial moles of \( PCl_3 \) = 0 mol Initial moles of \( Cl_2 \) = 0 mol Volume of the flask = 1 L Temperature = 500 K At equilibrium, moles of \( Cl_2 \) formed = 0.
1 mol.
From the stoichiometry of the reaction, at equilibrium: Moles of \( PCl_3 \) formed = moles of \( Cl_2 \) formed = 0.
1 mol Moles of \( PCl_5 \) remaining = initial moles - moles decomposed = 1 - 0.
1 = 0.
9 mol Since the volume is 1 L, the equilibrium concentrations are: \( [PCl_5] = 0.
9 \) mol/L \( [PCl_3] = 0.
1 \) mol/L \( [Cl_2] = 0.
1 \) mol/L The equilibrium constant in terms of concentration \( K_c \) is: $$ K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{(0.
1)(0.
1)}{0.
9} = \frac{0.
01}{0.
9} = \frac{1}{90} $$ Now, we need to find \( K_p \).
The relationship between \( K_p \) and \( K_c \) is: $$ K_p = K_c (RT)^{\Delta n_g} $$ where \( \Delta n_g \) is the change in the number of moles of gaseous species in the reaction.
\( \Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants}) \) \( \Delta n_g = (1 + 1) - 1 = 1 \) Given \( R = 0.
082 \) L atm mol\(^{-1} \) K\(^{-1} \) and \( T = 500 \) K.
$$ K_p = \left( \frac{1}{90} \right) (0.
082 \times 500)^1 = \frac{1}{90} \times 41 = \frac{41}{90} $$ $$ K_p \approx 0.
4555 $$ Rounding to three decimal places, \( K_p = 0.
456 \) atm.
The closest option is 0.
455 atm.