Question

The reaction mixture, at a certain temperature, contained concentrations of0.31 M of NH3, 0.85 M of N2 and 0.031 M of H2 when it reached equilibrium.Calculate Keq at this temperature.

Answer 1

Keq is 3798.42 or 3.798 x 10^3

Step-by-step explanation

Given that;

The concentration of NH3 is 0.31M

The concentration of N2 is 0.85M

The concentration of H2 is 0.031M

Follow the steps below to find the chemical equilibrium constant of the reaction

Step 1; Write the balanced equation of the reaction

`\text{ N}_(2(g))\text{ + 3H}_(2(g))\text{ }\rightleftarrows\text{ 2NH}_(3(g))`

Step 2; Write the chemical equilibrium constant for the reaction

`\begin{gathered} \text{ K}_(eq)\text{ = }\frac{K_f}{\text{ K}_b} \\ \text{ Where} \\ \text{ K}_f\text{ = }\lbrack NH_3\rbrack^2 \\ \text{ K}_b\text{ = }\lbrack N_2\rbrack\text{ }\lbrack H_2\rbrack^3 \end{gathered}`

Step 3; Substitute the kf and kb into the formula above

`\text{ K}_(eq)\text{ = }(\lbrack NH_3\rbrack^2)/(\lbrack H_2\rbrack^3*\lbrack N_2\rbrack)`

Recall, that

[NH3] = 0.31M

[H2] = 0.031M

[N2] = 0.85M

`\begin{gathered} \text{ K}_(eq)\text{ = }\frac{(0.31)^2}{(0.85)*\text{ \lparen0.031\rparen}^3} \\ \\ \text{ K}_(eq)\text{ = }\frac{\text{ 0.0961}}{0.85\text{ }*\text{ 0.0000298}} \\ \\ \text{ K}_(eq)\text{ = }\frac{\text{ 0.0961}}{0.0000253} \\ \\ \text{ K}_(eq)\text{ = 3798.42 or 3.798 }*\text{ 10}^3 \end{gathered}`

Therefore, Keq is 3798.42 or 3.798 x 10^3