Given the following chemical equilibria, $\begin{array}{l}
\begin{array}{lll}
\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) \
\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) \
\mathrm{H}_{2}(g)+1 / 2 \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)
\end{array}
\begin{array}{lll}
K_{1} \
K_{2} \
K_{3} \
\end{array}
\end{array}$

Determine the method used to calculate the equilibrium constant for the reaction below.
$4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$ $ \quad$$ \quad$$ \quad$ $K_{\mathrm{c}}$

Question

Given the following chemical equilibria, $\begin{array}{l}
\begin{array}{lll}
\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) \
\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) \
\mathrm{H}_{2}(g)+1 / 2 \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)
\end{array}
\begin{array}{lll}
K_{1} \
K_{2} \
K_{3} \
\end{array}
\end{array}$

Determine the method used to calculate the equilibrium constant for the reaction below.
$4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$  $ \quad$$ \quad$$ \quad$  $K_{\mathrm{c}}$

Quizplus · Accepted Answer

The equilibrium constant for the given reaction can be calculated by multiplying the equilibrium constants for the three reactions given in the problem statement in such a way that the products are on the right-hand side of the chemical equation and the reactants are on the left-hand side.

If we multiply the first reaction by two and reverse it, we get:

$$ 2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_2(g) + \mathrm{O}_2(g) \quad K_1^{-1} $$

Multiplying the second reaction by 2 and reversing it results in:

$$ 2 \mathrm{NH}_3(g) \rightleftharpoons \mathrm{N}_2(g) + 3 \mathrm{H}_2(g) \quad K_2^{-1} $$

Multiplying the third reaction by 2 and leaving it as it is, we get:

$$ \mathrm{H}_2(g) + \mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{H}_2 \mathrm{O}(g) \quad K_3 $$

Now, if we add these three reactions, we get:

$$ 2 \mathrm{NO}(g) + 2 \mathrm{NH}_3(g) + \mathrm{H}_2(g) + \mathrm{O}_2(g) \rightleftharpoons 3 \mathrm{N}_2(g) + 6 \mathrm{H}_2 \mathrm{O}(g) $$

The equilibrium constant for this reaction can be calculated by using the product of the equilibrium constants of the three reactions with the reactants and products raised to their stoichiometric coefficients, as given by the law of mass action.

$$ K_c = \frac{[\mathrm{NO}]^4 [\mathrm{H}_2\mathrm{O}]^6}{[\mathrm{NH}_3]^4 [\mathrm{O}_2]^5} = \frac{K_1^{-1} K_2^{-1} K_3}{[\mathrm{O}_2]^5} $$

Substituting the values of the equilibrium constants and simplifying, we get:

$$ K_c = \frac{K_1^2 K_3^6}{K_2^2} $$

Therefore, the correct choice is B.

Given the Following Chemical Equilibria Determine the Method Used to Calculate the Equilibrium Constant for \mathrm{H}_{2}(g)