For the reaction c(s) + co2(g) 2co(g) kc = 168. a mixture contains some c(s), [co] = 0.50 m and [co2] = 0.75 m. therefore the system at equilibrium, because

a. is not; the value of q is 0.67
b. is not; the value of q is 1.5
c. is; the value of q is 0.67
d. is not; the value of q is 0.33
e. is; the value of q is 0.33

Option D is correct. The system is not at equilibrium. Q = 0.33

Explanation:

Step 1: Data given

C(s) + CO2(g) ⇆ 2CO(g)       Kc = 168

[CO] = 0.50 M

[CO2] = 0.75 M

When Q=K, the system is at equilibrium and there is no shift to either the left or the right.

When Q<K, there are more reactants than products.

When Q>K, there are more products than reactants.

Step 2: Calculate Q

Q = [CO]² / [CO2]

Q = 0.50² / 0.75

Q = 0.33

Q < Kc

When Q<K, there are more reactants than products.

In order to reach equilibrium, the reaction will favor the forward reaction and try to use up some of the excess reactant to make more product.

Option D is correct

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