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Dmitry [639]
3 years ago
5

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
Chemistry
1 answer:
777dan777 [17]3 years ago
4 0

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

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