Question

A student ran the following reaction in the laboratory at 610 K:

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

When she introduced 0.183 moles of CO(g) and 0.211 moles of Cl2(g) into a 1.00 liter container, she found the equilibrium concentration of Cl2(g) to be 6.72×10-2 M.  

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

2.A student ran the following reaction in the laboratory at 546 K:

COCl2(g) CO(g) + Cl2(g)  

When she introduced 0.854 moles of COCl2(g) into a 1.00 liter container, she found the equilibrium concentration of COCl2(g) to be 0.817 M.  

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

3. 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

Answer 1

Part 1)

Initial concentration of Cl₂ = 0.211M

Initial concentration of CO = 0.183M

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

Initial--------------(0.183)---(0.211)---------(0)             

Equilibrium------(0.0392)--(0.0672)--------(0.1438)

Kc = [COCl₂]/[CO][Cl₂] = 0.1438/(0.0392 x 0.0672)   = 54.6

Part 2)

Initial concentration of COCl₂ = 0.854 M

--------------------COCl2(g) COCl₂(g) + Cl₂(g)

Initial--------------(0.854)--------------(0)---------(0)             

Equilibrium------(0.817)--------------(0.037)---(0.037)

Kc   = [CO][Cl₂]/[COCl₂] = (0.037 x 0.037)/0.817= 0.00167

Part 3)

Equilibrium concentration of COCl₂ = (6.56 x 10^-3 moles)/16.9 L = 3.88 x 10^-4 M

Equilibrium concentration of CO= (0.377)/16.9 L = 2.23 x 10^-2 M

Equilibrium concentration of Cl₂ = (0.372 moles)/16.9 L = 2.20 x 10^-2 M

Kc   = [CO][Cl₂]/[COCl₂] = [(2.20 x 10^-2 M)( 2.23 x 10^-2 M)]/ 3.88 x 10^-4 M = 1.2644

Kp = Kc(RT)

Here, R =0.0821 L atm /mol K

n = 2-1 = 1

Kp = 1.2644 x 0.0821 x 772 = 80.13

Kp = 80.13