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? \[ R = 0.082 { L atm mol}^{-1} {K}^{-1} \]
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? \[ R = 0.082 { L atm mol}^{-1} {K}^{-1} \]
Show Hint
- 7.38
- 73.8
- 0.738
- 0.043
The Correct Option is B
Solution and Explanation
\[ K_p = K_c(RT)^{\Delta n} \]
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 \( T = 500 { K} \):
\[ K_p = 1.8 \times (0.082 \times 500)^1 = 1.8 \times 41 = 73.8 { atm} \]
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