Question

At equilibrium, the concentrations of reactants and products can be predicted using the equilibrium constant, Kc, which is a mathematical expression based on the chemical equation. For example, in the reaction

aA+bB⇌cC+dD

where a, b, c, and d are the stoichiometric coefficients, the equilibrium constant is

Kc=[C]c[D]d[A]a[B]b

where [A], [B], [C], and [D] are the equilibrium concentrations. If the reaction is not at equilibrium, the quantity can still be calculated, but it is called the reaction quotient, Qc, instead of the equilibrium constant, Kc.

Qc=[C]tc[D]td[A]ta[B]tb

where each concentration is measured at some arbitrary time t.

a.)A mixture initially contains A, B, and C in the following concentrations: [A] = 0.700 M , [B] = 0.850 M, and [C] = 0.500 M . The following reaction occurs and equilibrium is established:

A+2B⇌C

At equilibrium, [A] = 0.570 M and [C] = 0.630 M . Calculate the value of the equilibrium constant, Kc.

Answer 1

Constructing ICE table for the reaction-

A + 2B <----> C

Initial(M): 0.700 0.850 0.500

Change: -x -2x +x

Eqlbrium: 0.700-x 0.850-2x 0.500+x

Now given,

Equilibrium concentrations,

[A]=0.570M

0.570M = 0.700M-x

x = 0.700M-0.500M = 0.130M

[C]= 0.630M

0.630M = 0.500M+x

x = 0.130M

Now keeping the value of x in [B]

[B]= 0.850-2x

[B] = 0.850-2*0.130 = 0.850M-0.260M

[B]=0.590M

Now that we know all the equilibrium concentrations-

Kc = [C]/[A][B]²

= 0.630M/0.570M*(0.590M)²

= 0.630M/0.19842M

Kc = 3.18