Question

What is the equilibriumC(s) + O₂(g) — CO₂(g)expression for the reaction below?

Answer 1

`\text{ Kc = }(\lbrack CO_2\rbrack)/(\lbrack O_2\rbrack)`

Step-by-step explanation

Given that

`\text{ C}_((s))\text{ + O}_(2(g))\text{ }\rightleftarrows\text{ CO}_(2(g))`

Follow the steps below to write the equilibrium expression for the reaction

`\begin{gathered} \text{ Kc}=\text{ }(kf)/(kb) \\ \text{ where kf is the rate of forward reaction} \\ \text{ kb is the rate of back ward reaction} \end{gathered}`

When writing equilibrium expression, we don't include substances that exist in solid state.

Hence, we have

`\begin{gathered} \text{ Kc = }\frac{\text{ kf}}{\text{ kb}} \\ \\ \text{ kf = }\lbrack\text{ CO}_2\text{ }\rbrack \\ \\ \text{ Kb = }\lbrack\text{ O}_2\text{ }\rbrack \\ \\ \text{ Hence, } \\ \text{ Kc = }\frac{\text{ }\lbrack\text{ CO}_2\text{ }\rbrack}{\lbrack O_2\rbrack} \end{gathered}`

Therefore, the correct option is B