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Given the equilibrium constant (K) which side of the reaction is favoured N2(g) + 3H2(g)  2NH3(g)        Keq =  735

Question

Given the equilibrium constant (K) which side of the reaction is favoured

N2(g)+3H2(g)2NH3(g) N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)
Keq=735 K_{eq} = 735

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Solution

The equilibrium constant (K) for a reaction is a measure of the ratio of the concentrations of the products to the reactants at equilibrium. If K > 1, the reaction is product-favored; if K < 1, the reaction is reactant-favored.

In this case, the equilibrium constant (Keq) for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g) is given as 735. Since this value is greater than 1, we can say that the reaction is product-favored. This means that at equilibrium, there will be more NH3 (the product) than N2 and H2 (the reactants).

This problem has been solved

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