Answer to Question #159094 in General Chemistry for Amr

Question

A mixture of 1 mol of N2 gas and 5 mol of H2 gas is introduced into a reactor of constant volume maintained at a constant temperature. Chemical equilibrium takes place the equation of which i:

N2 + 3H2 change 2NH3

THE MIXTURE CONTAINS AT THE FINAL STATE 0.65 MOLE OF AMMONIA GAS

1. Determine the molar composition of the mixture obtained at equilibrium

2. Did use the coefficient of transformation of gas H2 at equilibrium

3. calculate the equilibrium constant KC associated to this reaction knowing that the volume of the reactor is V=5L

Answer 1

N2 + 3 H2 = 2 NH3

1. Molar composition of the mixture obtained at equilibrium is:

n(NH3) = 0.65 mole;

n(N2) = 1 - 0.65/2 = 0.675 mole;

n(H2) = 5 - 0.65*3/2 = 4.025 mole;

2. Coefficient of transformation of H2 gas is:

n(H2) reacted / n(H2) initial = (5 - 4.025)/5 = 0.195

3. Equilibrium constant is:

Kс = С(NH3)²/{C(N2)*C(H2)³}, where С = n/V (mole/L)

Kc = (0.65/5)²/{(0.675/5)*(4.025/5)³} = 0.240