1. A student ran the following reaction in the laboratory at 590. K: CO(g) + Cl2

Question

1. A student ran the following reaction in the laboratory at 590. K:

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

When she introduced CO(g) and Cl2(g) into a 1.00 L evacuated container, so that the initial partial pressure of CO was 2.81 atm and the initial partial pressure of Cl2 was 1.87 atm, she found that the equilibrium partial pressure of COCl2 was 1.42 atm.

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

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.

4The equilibrium constant, Kc, for the following reaction is 9.52×10-2 at 350 K.

CH4 (g) + CCl4 (g) 2 CH2Cl2 (g)

Calculate the equilibrium concentrations of reactants and product when 0.375 moles of CH4 and 0.375 moles of CCl4 are introduced into a 1.00 L vessel at 350 K.

Explanation / Answer

Lets solve question No. 1 , This question is related to Equilibrium constant in terms of partial pressure

Now we have

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

Initial 2.81 1.87 0

Final 2.81-x 1.87-x x=1.42

Final 1.39 0.45 1.42

Now we know

So

Kp = (pCOCl2 ) / pCO * pCl2 =   1.42 / (1.39 * 0.45) =2.27 /atm

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