The equilibrium constant, Kp, for the following reaction is 0.497 at 500. K. PC15 (9) PC13 (g) + Cl₂ (g) If an equilibrium mixture of the three gases in a 12.7 L container at 500. K contains PC15 at a pressure of 0.804 atm and PC13 at a pressure of 0.911 atm, the equilibrium partial pressure of Cl₂ is l atm.

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**Equilibrium Constant Problem**

**Reaction and Conditions:**

The equilibrium constant, \( K_p \), for the following reaction is 0.497 at 500 K.

\[ \text{PCl}_5 (g) \rightleftharpoons \text{PCl}_3 (g) + \text{Cl}_2 (g) \]

**Problem Statement:**

If an equilibrium mixture of the three gases in a 12.7 L container at 500 K contains:

- \(\text{PCl}_5\) at a pressure of 0.804 atm
- \(\text{PCl}_3\) at a pressure of 0.911 atm

Determine the equilibrium partial pressure of \(\text{Cl}_2\).

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Transcribed Image Text:**Equilibrium Constant Problem** **Reaction and Conditions:** The equilibrium constant, \( K_p \), for the following reaction is 0.497 at 500 K. \[ \text{PCl}_5 (g) \rightleftharpoons \text{PCl}_3 (g) + \text{Cl}_2 (g) \] **Problem Statement:** If an equilibrium mixture of the three gases in a 12.7 L container at 500 K contains: - \(\text{PCl}_5\) at a pressure of 0.804 atm - \(\text{PCl}_3\) at a pressure of 0.911 atm Determine the equilibrium partial pressure of \(\text{Cl}_2\). --- **Interactive Elements:** - **Submit Answer**: Button to submit your answer. - **Retry Entire Group**: Option to retry if the solution attempt is incorrect. There are 9 more group attempts remaining.
Expert Solution
Step 1: Interpretation of datas :

G i v e n space colon
R e a c t i o n space colon space P C I subscript 5 left parenthesis g right parenthesis space rightwards harpoon over leftwards harpoon P C I subscript 3 left parenthesis g right parenthesis space plus C I subscript 2 left parenthesis g right parenthesis
V o l u m e space o f space f l a s k space equals space 12.7 space L
T e m p e r a t u r e comma space T space equals space 500 space K
E q u i l i b r i u m space c o n c e n t r a t i o n space s colon
P subscript P C I subscript 5 end subscript equals 0.804 space a t m
P subscript P C I subscript 3 end subscript equals 0.911 space a t m
P subscript C I subscript 2 end subscript equals ?
K subscript p equals 0.497 space
O b j c e t i v e space colon
C a l c u l t a e space e q u i l i b r i u m space p r e s s u r e space o f space C I subscript 2

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