Write the expression for \(K_{\text {eq }}\) for the reaction \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightleftarrows 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\)

Short Answer

Expert verified
The expression for the equilibrium constant, \(K_{eq}\), for the given reaction is: \[K_{eq} = \frac{[NO]^4[H_2O]^6}{[NH_3]^4[O_2]^5}\]

Step by step solution

01

Identify the reactants and products

In the given reaction, the reactants are: - Ammonia (NH3) with a coefficient of 4 - Oxygen (O2) with a coefficient of 5 The products are: - Nitric oxide (NO) with a coefficient of 4 - Water vapor (H2O) with a coefficient of 6
02

Write the expression for the equilibrium constant \(K_{eq}\)

Using the general formula for the equilibrium constant, we have \[K_{eq} = \frac{[NO]^4[H_2O]^6}{[NH_3]^4[O_2]^5}\] This gives us the expression for the equilibrium constant for the given reaction.

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