Question

At 500 K, for the reaction \[ {PCl}_5(g) \leftrightarrow {PCl}_3(g) + {Cl}_2(g) \] the equilibrium constant \( K_c \) is 1.8 mol/L. What is \( K_p \) in atm at the same temperature? \[ R = 0.082 { L atm mol}^{-1} {K}^{-1} \] 

• A. 7.38
• B. 73.8
• C. 0.738
• D. 0.043

Hint:

When converting \( K_c \) to \( K_p \), remember to factor in the temperature in Kelvin and the universal gas constant in the correct units to ensure consistency across the calculation.

Answer 1

The Correct Option is B

The relationship between \( K_c \) and \( K_p \) for the reaction can be derived using the equation:
\[ K_p = K_c(RT)^{\Delta n} \]
where \( \Delta n \) is the change in moles of gas (products - reactants). For this reaction:
\[ \Delta n = (1 + 1) - 1 = 1 \]
Given that \( R = 0.082 { L atm mol}^{-1} {K}^{-1} \) and \( T = 500 { K} \):
\[ K_p = 1.8 \times (0.082 \times 500)^1 = 1.8 \times 41 = 73.8 { atm} \]