Question

Here's a specific problem for the student to solve:Consider the reaction A(g) + B(g) ↔ C(g) with a standard enthalpy change of reaction (ΔH°rxn) of -50 kJ/mol at 298 K. If the initial concentrations of A and B are both 1.0 M, what will be the equilibrium concentration of each gas at 298 K? The equilibrium constant (Kc) for the reaction at 298 K is 2.75 x 10^-2.

Answer 1

To solve this problem, we will use the equilibrium constant expression and an ICE  Initial, Change, Equilibrium  table.The balanced equation for the reaction is:A g  + B g   C g The equilibrium constant expression for this reaction is:Kc = [C] /  [A] * [B] Given that Kc = 2.75 x 10^-2 at 298 K, we can set up the ICE table:Initial concentrations:[A] = 1.0 M[B] = 1.0 M[C] = 0 MChange in concentrations:[A] = -x[B] = -x[C] = +xEquilibrium concentrations:[A] = 1.0 - x[B] = 1.0 - x[C] = xNow, we can plug the equilibrium concentrations into the equilibrium constant expression:2.75 x 10^-2 = x /   1.0 - x  *  1.0 - x  To solve for x, we can first simplify the denominator:2.75 x 10^-2 = x /  1 - 2x + x^2 Now, multiply both sides by the denominator to get rid of the fraction:x = 2.75 x 10^-2 *  1 - 2x + x^2 Since Kc is small, we can assume that x is also small. Therefore, we can approximate 1 - 2x  1:x  2.75 x 10^-2 *  1 - 0 + 0 x  2.75 x 10^-2Now, we can find the equilibrium concentrations of each gas:[A] = 1.0 - x  1.0 - 2.75 x 10^-2  0.975 M[B] = 1.0 - x  1.0 - 2.75 x 10^-2  0.975 M[C] = x  2.75 x 10^-2 MSo, the equilibrium concentrations of A, B, and C at 298 K are approximately 0.975 M, 0.975 M, and 2.75 x 10^-2 M, respectively.