Consider the following reaction:N2 + 3H2 2NH3i. Derive expression of Kc for the above reactionii. Calculate equilibrium concentration of N2. The equilibrium concentration of H2and NH3 are 1.0 moldm3 and 0.5 moldm-3 respectively. Kc of above reaction at25℃ is 1.85 ×10-3
Question
Consider the following reaction:N2 + 3H2 2NH3i. Derive expression of Kc for the above reactionii. Calculate equilibrium concentration of N2. The equilibrium concentration of H2and NH3 are 1.0 moldm3 and 0.5 moldm-3 respectively. Kc of above reaction at25℃ is 1.85 ×10-3
Solution
i. Derive expression of Kc for the above reaction
The reaction is: N2 + 3H2 ⇌ 2NH3
The equilibrium constant expression (Kc) for a reaction is the ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of its stoichiometric coefficient in the balanced chemical equation.
So for the above reaction, the expression for Kc would be:
Kc = [NH3]^2 / ([N2] * [H2]^3)
ii. Calculate equilibrium concentration of N2
We are given that the equilibrium concentrations of H2 and NH3 are 1.0 moldm-3 and 0.5 moldm-3 respectively, and that Kc is 1.85 ×10-3.
Substituting these values into the Kc expression, we get:
1.85 ×10-3 = (0.5)^2 / ([N2] * (1.0)^3)
Solving for [N2], we get:
[N2] = (0.5)^2 / (1.85 ×10-3 * 1.0)
[N2] = 0.25 / 1.85 ×10-3
[N2] = 0.135 moldm-3
So the equilibrium concentration of N2 is approximately 0.135 moldm-3.
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