2CO 2(g) ↔ 2CO (g) + O 2(g)

Question

0.30 mol of CO 2 is placed in 2.3L reaction vessel and is heated to 2500 ̊C. What would you expect the equilibrium concentration of each entity to be if K for the above reaction is 8.30 x 10 -7 at 2500 ̊C?

Answer 1

This is an equilibrium concentration problem. Given the equation, the K expression is [CO]²[O₂]/[CO]²

If x is the CO₂ change in molarity (molarity can be a surrogate for moles at constant V) then

8.3 x 10^-7 = (2x)²(x)/(.3-2x)²

We can assume that 2x << .3 and solve 8.3 x 10^-7 = 4x³/.3 The approximation satisfies 5 % rule.

Answer 2

This problem is much easier to solve with an ICE box, but that's hard to format here.

2CO2(g)<-->2CO(g)+O2(g)

[CO2]initial=0.3mol/2.3L=0.1304 M

[CO]initial=0

[O2]initial=0

[CO2] decreases by 2x

[CO] increases by 2x

[O2] increases by x

write the K expression

K=([CO]²[O₂]) / ([CO₂]²)

substitute the K, and combine the initial and the change expressions:

8.3(10^-7)=([2x]²[x]) / ([2.3-2x]²)

omit the -2x for CO2 because it's so small

8.3(10^-7)=([2x]²[x]) / ([2.3]²)

solve for x

x≈0.0103155

[CO2]eq=2.3-2x=2.3-2(0.0103155)=0.109769 M

[O2]eq=x=0.0103155 M

[CO]eq=2x=0.0103155(2)=0.020631 M