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For a particular reaction K = 98000. What can be said about this reaction?
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16. Chemical Equilibrium
Equilibrium Constant K
Multiple Choice
At T = 700 K, the equilibrium constant for the reaction CCl4(g) ⇌ C(s) + 2 Cl2(g) is K_p = 0.76. A flask is charged with 3.0 atm of CCl4(g) at 700 K, which is then allowed to reach equilibrium. What is the partial pressure of Cl2(g) in the flask?
A
1.52 atm
B
3.04 atm
C
0.76 atm
D
2.28 atm
Verified step by step guidance1
Identify the balanced chemical equation: CCl4(g) ⇌ C(s) + 2 Cl2(g). Note that C(s) is a solid and does not appear in the expression for the equilibrium constant K_p.
Write the expression for the equilibrium constant K_p in terms of partial pressures: K_p = (P_{Cl2}^2) / (P_{CCl4}).
Let x be the change in pressure of CCl4 that dissociates at equilibrium. Therefore, the change in pressure for Cl2 will be 2x, since 2 moles of Cl2 are produced for every mole of CCl4 that dissociates.
Set up the equilibrium expressions: P_{CCl4} = 3.0 - x and P_{Cl2} = 2x.
Substitute these expressions into the K_p expression: 0.76 = (2x)^2 / (3.0 - x). Solve this equation for x to find the equilibrium partial pressure of Cl2, which is 2x.
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