Question 15.12: Consider the reaction: N2( g) + 3 H2( g) ⇄ 2 NH3( g) Keq = 1......
Consider the reaction:
\mathrm{N}_{2}(g)\,+\,3\,\mathrm{H}_{2}(g)\,\rightleftarrows\,2\,\mathrm{NH}_{3}(g)K_{\mathrm{eq}}=152~\mathrm{at}~225^{\circ}\mathrm{C}
In an equilibrium mixture, [N_2] = 0.110 M and [H_2] = 0.0935 M. What is the equilibrium concentration of NH_3?
GIVEN: [N_2] = 0.110 M
[H_2] = 0.0935 M
K_{eq} = 152
FIND: [NH_3]
SOLUTION MAP
[N_2], [H_2], K_{eq} → [NH_3]
K_{\mathrm{eq}}={\frac{\mathrm{[NH_{3}]}^{2}}{\mathrm{[N_{2}]}[\mathrm{H}_{2}]^{3}}}
Step-by-Step
K_{\mathrm{eq}}={\frac{\mathrm{[NH_{3}]}^{2}}{\mathrm{[N_{2}]}[\mathrm{H}_{2}]^{3}}}
[{ N}{ H}_{3}]^{2}=K_{\mathrm{eq}}[{ N}_{2}][{ H}_{2}]^{3}
[{\mathrm{NH}}_{3}]={\sqrt{K_{\mathrm{eq}}[{\mathrm{N}}_{2}][{\mathrm{H}}_{2}]^{3}}}
={\sqrt{(152)(0.110)(0.0935)^{3}}}
= 0.117 M
K_{\mathrm{eq}}={\frac{\mathrm{[NH_{3}]}^{2}}{\mathrm{[N_{2}]}[\mathrm{H}_{2}]^{3}}} = \frac{(0.117)^{2}}{(0.110)(0.0935)^{3}} = 141
The calculated value of K_{eq} is about equal to the given value of K_{eq} (which was 152), indicating that your answer is correct. The slight difference is due to rounding, which is common in problems like these.
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