Question

The table below gives the equilibrium concentrations for this reaction at acertain temperature:N₂(g) + O₂(g) →→→→2NO(g)0.69 M[N₂]0.98 MOA. 20OB. 1.7 x 10-3OC. 5.0 × 10-2OD. 9.9[0₂]What is the equilibrium constant for the reaction?0.034 M[NO]

Answer 1

Option B is correct

`1.7*10^(-3)`

Explanations:

Given the chemical reaction below;

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

The equilibrium constant for the reaction is given as;

`k=([NO]^2)/([O_2][N_2])`

Given the following parameters

`\begin{gathered} [NO]=0.034M \\ [N_2]=0.69M \\ [O_2]=0.98M \end{gathered}`

Substitute

`\begin{gathered} k=((0.034)^2)/((0.69)(0.98)) \\ k=(0.001156)/(0.6762) \\ k=0.001709=1.7*10^(-3) \end{gathered}`

Therefore the equilibrium constant for the reaction is 1.7 * 10^-3