Question

Part A A mixture initially contains A, B, and in the following concentrations: [A] = 0.500 M, [B] = 1.05 M, and [C] = 0.350 M. The following reaction occurs and equilibrium is established: A +2B=C At equilibrium, [A] = 0.320 M and [C] = 0.530 M Calculate the value of the equilibrium constant, K. Express your answer numerically. View Available Hint(s)

Answer 1

The given reaction is

A + 2B $\rightleftharpoons$ C

As per law of equilibrium at equilibrium position the ratio of product of concentrations of products to the product of concentrations of reactants remains constant. This is called as equilibrium constant. Represented with K_{​​​​​​C}

Equilibrium constant (K_{​​​​​​C}) = [C] /[A] [B]²

Here the concentrations of the reactants and products at equilibrium position.

Initial concentrations of [A] = 0.500 M

[B] = 1.05 M [C] = 0.350 M

Note: As per balanced equation if reactant A is consumed by X Molar concentration the reactant B will be consumed by 2X Molar concentration and the reactant C will be formed by X Molar concentration.

Given that equilibrium concentration of [A] = 0.320 M

The concentration of A consumed = 0.180 M

Then the concentration of B at equilibrium position = 1.05 M - 2 (0.180 M) = 0.690 M

The equilibrium concentration of C = 0.530 M (0.350 M + 0.180 M)

Now equilibrium constant

Kc = (0.530 M)/(0.320 M) (0.690 M)²

= 0.530 M / 0.15235 M^{​​​​​​3}

= 3.47878 M^{​​​​​​-2}

Hence the equilibrium constant Kc = 3.48 M^{​​​​​​-2}