Question

A chemist mixes 0.1 mol of hydrogen (H2) with 0.2 mol of nitrogen (N2) in a 2 L flask at 400°C. The reaction H2(g) + N2(g) ⇌ 2NH3(g) establishes equilibrium. Calculate the equilibrium concentrations of hydrogen, nitrogen, and ammonia if the Kc value of the reaction at this temperature is 1.5 x 10^-2.

Answer 1

To calculate the equilibrium concentrations of hydrogen, nitrogen, and ammonia, we will use the initial concentrations and the Kc value given.Initial concentrations:[H2] = 0.1 mol / 2 L = 0.05 M[N2] = 0.2 mol / 2 L = 0.1 M[NH3] = 0 since no ammonia is present initiallyLet x be the change in concentration at equilibrium:[H2] = 0.05 - x[N2] = 0.1 - x[NH3] = 2xNow, we can write the expression for Kc:Kc = [NH3]^2 /  [H2] * [N2] Plug in the equilibrium concentrations and the given Kc value:1.5 x 10^-2 =  2x ^2 /   0.05 - x  *  0.1 - x  Now, we need to solve for x: 1.5 x 10^-2  *   0.05 - x  *  0.1 - x   =  2x ^2Solving for x, we get:x  0.014Now, we can find the equilibrium concentrations:[H2] = 0.05 - x  0.05 - 0.014  0.036 M[N2] = 0.1 - x  0.1 - 0.014  0.086 M[NH3] = 2x  2 * 0.014  0.028 MSo, the equilibrium concentrations are approximately:[H2]  0.036 M[N2]  0.086 M[NH3]  0.028 M