Question

has a standard free-energy change of –3.82 kJ/mol at 25 °C. What are the concentrations of A, B, and C at equilibrium if, at the beginning of the reaction, their concentrations are 0.30 M, 0.40 M, and 0 M, respectively

Answer 1

Solution :-

Delta G= -3.82 kJ *1000 J / 1kJ = -3820 J

T = 25 + 273 = 298 K

A + B ----- > C

Using the delta G lets calculate the equilibrium constant K

Delta G = - RT ln K

Delta G / -R*T= ln K

Antil n (delta G / -RT) = K

0.214 = K

Now lets make the ICE table

  A     +       B        ----- > C

0.30 M    0.40 M         0

-x              -x                +x

0.30-x      0.40-x         x

K = [C]/[A][B]

0.214 = [x]/[0.30-x][0.40-x]

0.214 * [0.30-x][0.40-x] = x

Solviong for the x we get

0.0224 = x

So the equilibrium concentrations are as follows

[A] = 0.30- x = 0.30 M – 0.0224 M = 0.278 M

[B] =0.40-x = 0.40 M – 0.0224 M = 0.0378 M

[C] = 0.0224 M