The equilibrium constant, \(K _{ C }\) for the reaction, \(NO _{( g )}+\) \(1 / 2 O _{2(\beta)} \rightarrow NO _{2( g )}\), is \(35.2\). What is the value of \(K _{ C }\) for the reaction, \(NO _{2( g )} \rightarrow NO _{(\varepsilon)}+1 / 2 O _{2(\varepsilon)}\) ?
A. \(35.2\) B. \(17.6 \) C. \(2.84 \times 10^{-2}\) D. \(1.24 \times 10^{3}\)
Correct Answer: C
Explanation
\begin{aligned} & NO +1 / 2 O _{2} \rightarrow NO _{2(g)}, k _{ c }=35.2 \\ & NO _{2} \rightarrow NO +1 / 2 O _{2}, k _{ c }=? \end{aligned} The equilibrium constant of a reversed reaction is a multiplicative inverse of that of the formed reaction i.e. \(k_{c \text { coack- }}\) ward) $$ =\frac{1}{k_{c(f \text { funurid) }}}=\frac{1}{35.2} \Rightarrow 2.84 \times 10^{-2} $$
