For the reaction N_1(g) + 3H_2 2NH_3 (g) what is the value of K_c at 500 degreeC

Question

For the reaction N_1(g) + 3H_2 2NH_3 (g) what is the value of K_c at 500 degreeC if the equilibrium concentrations are as follows: [H_2] = 0.35 M, [N_2] = 0.40 M, and [NH_3] = 1.9 M? This reaction favors the formation of the product. NH3, so the equilibrium constant should be greater than 1. Set up the equilibrium-constant expression for this reaction using the equilibrium concentrations. The product concentrations are in the numerator, and the reactant concentrations are in the denominator. Additionally, the coefficients from the balanced chemical equation become exponents for the respective concentrations.

Explanation / Answer

K = NH3^2/(N2)(H2^3)

K = ?

when

equilibrium concnentration

NH3=1.9

N2 = 0.40

H2 = 0.35

then

K = NH3^2/(N2)(H2^3)

K = 1.9 ^2/((0.40)(0.35^3))

K = 210.495

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