In the reaction N_{2(g)} + 3H_{2(g)} rightlef | vedclass

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(C) The given reaction is $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ with $\Delta H = -93.6 \ kJ$. Since $\Delta H < 0$,the reaction is exot

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Pressure is lowered

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(C) The given reaction is $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ with $\Delta H = -93.6 \ kJ$. Since $\Delta H The total number of moles of gaseous reactants is $1 + 3 = 4$,and the number of moles of gaseous products is $2$. Increasing the pressure or decreasing the volume of the reaction vessel shifts the equilibrium toward the side with fewer moles,which is the product side,thereby increasing the yield of $NH_3$. Conversely,lowering the pressure shifts the equilibrium toward the side with more moles (the reactant side),which decreases the yield of $NH_3$.

In the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}; \Delta H = -93.6 \ kJ$,the yield of ammonia does not increase when:

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