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
