Question

tionThe table below gives the equilibrium concentrations for this reaction at acertain temperature:N (9) +0,0).22NO(9)[N][02][NO]0.69 M0.98 M0.034 MWhat is the equilibrium constant for the reaction?A. 9.9B. 1.7 x 10-3

Answer 1

Answer:

k = 1.7 x 10^-3

Explanations:

Given the the reaction between nitrogen and oxygen gas expressed as:

$N_2(g)+O_2(g)\rightarrow2NO(g)$

The equilibrium constant for the reaction is expressed as:

$k=\frac{[NO]^2}{[N_2[O_2]}$

Given the following concentrations

[NO] = 0.034M

[N2] = 0.69

[O2] = 0.98

Substitute the concentrations into the formula to have:

$\begin{gathered} k=\frac{0.034^2}{0.69\times0.98} \\ k=\frac{0.001156}{0.6762} \\ k=0.0017 \\ k=1.7\times10^{-3} \end{gathered}$

Hence the equilibrium constant for the reaction is 1.7 x 10^-3