Consider the following reaction: COCl2(g) CO(g) + Cl2(g) If 6.56×10-3 moles of C

Question

Consider the following reaction: COCl2(g) CO(g) + Cl2(g) If 6.56×10-3 moles of COCl2, 0.377 moles of CO, and 0.372 moles of Cl2 are at equilibrium in a 16.9 L container at 772 K, the value of the equilibrium constant, Kp, is

2.A student ran the following reaction in the laboratory at 597 K:

COCl2(g) CO(g) + Cl2(g)

When she introduced COCl2(g) at a pressure of 0.710 atm into a 1.00 L evacuated container, she found the equilibrium partial pressure of COCl2(g) to be 0.299 atm.  

Calculate the equilibrium constant, Kp, she obtained for this reaction.  

Explanation / Answer

1) for the following reaction:

COCl2(g) -------------> CO(g) + Cl2(g)

for this reaction , we have

Kc = [Cl2][CO]/[COCl2]

= ( 0.372 mol / 16.9 L ) *( 0.377 mol / 16.9 L ) / (6.56×10-3 mol / 16.9 L)

= 0.022 * 0.0223 / (0.388 x 10^-3)

= 1.27

and

Kp = Kc (RT)^deltan

= 1.27 * ( 8.314 J/K-mol x 772 K)^(2-1)

= 1.27 * ( 8.314 J/K-mol x 772 K)

= 8151.38

therefore

Kp = 8151.38

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