At high temps carbon reacts with O2 to produce as follows C(s) + O2(g) <> 2CO(g) when 0.350 mol of O2 and excess carbon were placed in a 5.00-L container and heated the equil. concentration of CO was found to be 0.060 M. What is the equil. constant, Kc, for this reaction?

(1) 0.010
(2) 0.072
(3) 1.2
(4) 0.17
(5) 0.090

(5) 0.090

Explanation:

The initial concentration of O₂ is:

0.350 mol / 5.00 L = 7.00 × 10⁻² mol/L

We can find the concentrations at equilibrium using an ICE chart.

      2 C(s)   +   O₂(g)   ⇄   2 CO(g)

I                  7.00 × 10⁻²         0

C                      -x                  +2x

E             7.00 × 10⁻² - x         2x

Since the concentration at equlibrium of CO was 0.060 M,

2x = 0.060 M

x = 0.030 M

The concentrations at equilibrium are:

[O₂] = 7.00 × 10⁻² - x = 7.00 × 10⁻² - 0.030 = 0.040 M

[CO] = 2x = 2 (0.030) = 0.060 M

The equilibrium constant (Kc) is:

Kc = [CO]²/[O₂]

Kc = (0.060)²/(0.040)

Kc = 0.090

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in chemistry, a solution’s concentration is how much of a dissolvable substance, known as a solute, is mixed with another substance, called the solvent. the standard formula is c = m/v, where c is the concentration, m is the mass of the solute dissolved, and v is the total volume