When 0.16 mol of O2 gas and excess graphite were placed in a 2.00-L container and come to equilibrium, the concentration of CO gas was 0.034 M

Question

When 0.16 mol of O2 gas and excess graphite were placed in a 2.00-L container and come to equilibrium, the concentration of CO gas was 0.034 M. What is the equilibrium constant, Kc, for this reaction at this temperature? Please report 4 decimal places, without units.

C(s) + O2(g)  2CO(g)

Answer 1

It's probably best to set up an ICE table for these types of problems.

C(s) + O₂(g) <==> 2CO(g)

.............0.16.............0.........Initial

.............-x...............+2x.......Change

--------0.16-x..........2x..........Equilibrium

Since at equilibrium we are told [CO] = 0.034 M and there are 2.0 liters, the moles of CO = 0.034 mol/L x 2 L =0.068 moles of CO at equilibrium. Thus 2x = 0.068 moles and x = 0.034 mols.

So @ equilibrium ....

[O₂] = 0.16 mols - 0.034 mols = 0.126 mols/2 L = 0.063 M

[CO] = 0.034 M

K = [CO]² / [O₂] = (0.034)² / 0.063

K = 0.0183

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To use M concentrations, we can proceed as follows:

C(s) + O₂(g) <==> 2CO(g)

......0.08...................0.........Initial

......-x.....................+2x.......Change

0.08-x...................2x.........Equilibrium

At equilibrium, 2x = 0.034 so x = 0.017 M

[O₂] = 0.08 - 0.017 = 0.063 M

[CO] = 0.034 M

K = [CO]2 / [O₂] = (0.034)² / 0.063

K = 0.0183

Answer 2

When 0.16 mol of O2 gas and excess graphite were placed in a 2.00-L container and come to equilibrium, the concentration of CO gas was 0.034 M. What is the equilibrium constant, Kc, for this reaction at this temperature? Please report 4 decimal places, without units.

C(s) + O2(g)  2CO(g)

Kc = [Products_eq]^coeff/[Reactants_eq]^coeff *Of the gaseous forms only, NO solid or liquid, so NO Carbon.

K_c = [CO_g]²/[O_2g]

K_c = [CO_g]²/[O_2g]

K_c = [CO_g + x]²/[O_2g - x] at equillibrium

Kc = [0.034]²/[0.08 - 0.034]

= .001156 / .046

K_c = 0.0251