The Haber process is commonly written as At 500 ,K

CHEMISTRY · Single correct · A Level · Chemistry · Chemical equilibria (Le Chatelier’s principle and Kc) · Chemical equilibria (Le Chatelier’s principle and Kc)

Problem

The Haber process is commonly written as

\text{N}_2(g) + 3\,\text{H}_2(g) \rightleftharpoons 2\,\text{NH}_3(g)

At 500\,K, the equilibrium constant for this reaction is

K = 6.0 \times 10^{-2}

. In a particular vessel, the instantaneous partial pressures give

Q_0 = 0.25

for this orientation. A chemist rewrites the equation by first multiplying all stoichiometric coefficients by 3 and then reversing the reaction, obtaining

6\,\text{NH}_3(g) \rightleftharpoons 3\,\text{N}_2(g) + 9\,\text{H}_2(g)

Using the same set of partial pressures, which of the following gives the correct value of the reaction quotient

Q_{\text{mod}}

for the modified reaction and predicts whether ammonia

(\text{NH}_3)

will be formed or consumed as the system moves toward equilibrium in this orientation? (a)

Q_{\text{mod}} = 64

and

Q_{\text{mod}} > K_{\text{mod}}

; ammonia will be formed. (b)

Q_{\text{mod}} = 64

and

Q_{\text{mod}} < K_{\text{mod}}

; ammonia will be consumed. (c)

Q_{\text{mod}} = 64

and

Q_{\text{mod}} < K_{\text{mod}}

; ammonia will be formed. (d)

Q_{\text{mod}} = 4

and

Q_{\text{mod}} < K_{\text{mod}}

; ammonia will be consumed.

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