Question

In the gas-phase reaction 2A + B <-> 3C + 2D, it was found that, when 1.00 mol A, 2.00 mol B, and 1.00 mol D were mixed and allowed to reach equilibrium at 25°C, the resulting mixture contained 0.90 mol C at a total pressure of 1.00 bar.
1. Calculate the mole fraction of each species at equilibrium, the equilibrium constant K and the standard reaction free energy change? .

Answer 1

Answer:

Mole fraction: A =  8.70%, B = 37.00%, C =  19.60%, D = 34.80%

K = 6.86

Standard reaction free energy change: -4.77 kJ/mol

Explanation:

Let's do an equilibrium chart for the reaction:

2A + B ⇄ 3C + 2D

1.00  2.00  0     1.00    Initial

-2x     -x     +3x   +2x    Reacts (stoichiometry is 2:1:3:2)

1-2x    2-x    3x    1+2x  Equilibrium

3x = 0.9

x = 0.3 mol

Thus, the number of moles of each one at the equilibrium is:

A = 1 - 2*0.3 = 0.4 mol

B = 2 - 0.3 = 1.7 mol

C = 0.9 mol

D = 1 + 2*0.3 = 1.6 mol

The molar fraction is the mol of the component divided by the total number of moles (0.4 + 1.7 + 0.9 + 1.6 = 4.6 mol):

A = 0.4/4.6 = 0.087 = 8.70%

B = 1.7/4.6 = 0.37 = 37.00%

C = 0.9/4.6 = 0.196 = 19.60%

D = 1.6/4.6 = 0.348 = 34.80%

The equilibrium constant is the multiplication of the concentration of the products elevated by their coefficients, divided by the multiplication of the concentration of the reactants elevated by their coefficients. Because the volume remains constant, we can use the number of moles:

K = (nC³*nD²)/(nA²*nB)

K = (0.9³ * 1.6²)/(0.4² * 1.7)

K = 6.86

The standard reaction free energy change can be calculated by:

ΔG° = -RTlnK

Where R is the gas constant (8.314 J/mol.K), and T is the temperature (25°C = 298 K)

ΔG° = -8.314*298*ln6.86

ΔG° = -4772.8 J/mol

ΔG° = -4.77 kJ/mol