Question

At high temperatures, carbon reacts with O2 to produce CO 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 equilibrium concentration of CO was found to be 0.060 M. What is the equilibrium constant, Kc, for this reaction

Answer 1

Answer:

Value of $K_{c}$ is 0.090.

Explanation:

Initial molarity of $O_{2}$ = $\frac{0.350}{5.00}M$ = 0.0700 M

Construct an ICE table corresponding to the combustion reaction of carbon to determine $K_{c}$

                       $C(s)+O_{2}(g)\rightarrow 2CO(g)$

              I (M):    -      0.0700        0

             C (M):  -          -x             +2x

             E (M):   -    0.0700-x       2x

So, $K_{c}=\frac{[CO]^{2}}{[O_{2}]}$  , where [CO] and $[O_{2}]$ represents equilibrium concentration of CO and $O_{2}$ respectively.

Here, $[CO]=2x=0.060$

       ⇒x = 0.030

So, $[O_{2}]$ = 0.0700-x = (0.0700-0.030) = 0.040

Hence,  $K_{c}=\frac{(0.060)^{2}}{0.040}=0.090$