Question

For the gas phase homogeneous equilibrium:

\[ 2X(g) \rightleftharpoons 2Y(g) + Z(g) \]

$K_C$ at 400K is $1 \times 10^{-3}$ mol L^-1. What is the value of $K_P$ for the equilibrium at 400K?

\[ R = 0.082 \, \text{L atm K}^{-1} \, \text{mol}^{-1} \]

• A. $1 \times 10^{-3}$ atm
• B. $3.16 \times 10^{-4}$ atm
• C. $4.24 \times 10^{-4}$ atm
• D. $3.28 \times 10^{-2}$ atm
• E. $1.28 \times 10^{-2}$ atm

Hint:

Use the formula $K_P = K_C (RT)^{\Delta n}$ to convert between equilibrium constants.

Answer 1

The Correct Option is D

Step 1: The relation between $K_P$ and $K_C$ is given by: \[ K_P = K_C (RT)^{\Delta n} \] where $\Delta n$ is the change in the number of moles of gaseous products and reactants. 
Step 2: From the reaction: \[ \Delta n = (2 + 1) - 2 = 1 \] Step 3: Given values: \[ K_C = 1 \times 10^{-3}, \quad R = 0.082, \quad T = 400 { K} \] Step 4: Calculating $K_P$: \[ K_P = (1 \times 10^{-3}) (0.082 \times 400)^1 \] \[ K_P = (1 \times 10^{-3}) (32.8) = 3.28 \times 10^{-2} { atm} \] Step 5: Therefore, the correct answer is (D).