Question

1 mole of nitrogen and 3 moles of hydrogen were mixed at 593k and 2.0x10^7pa.At equilibrium the mixture contain 1.5 mole of ammonia.
1. Write the equilibrium expression for Kc
2. Find the mole fraction of each component and thus their molar concentration.

Answer 1

1. The equation for the reaction is:

N2(g) + 3H2(g) ⇌ 2NH3(g)

The equilibrium expression for Kc is:

Kc = [NH3]^2 / [N2] [H2]^3

2. To find the mole fraction of each component, we first need to calculate the total number of moles in the mixture:

Total moles = 1 mole of N2 + 3 moles of H2 = 4 moles

The mole fraction of N2 is:

X(N2) = moles of N2 / total moles = 1 / 4 = 0.25

The mole fraction of H2 is:

X(H2) = moles of H2 / total moles = 3 / 4 = 0.75

To calculate the molar concentration of each component, we need to use the ideal gas law:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (8.31 J/mol/K), and T is the temperature (593 K).

For N2:

n(N2) = PV / RT = (2.0x10^7 Pa)(1.0 L) / (8.31 J/mol/K)(593 K) = 0.521 mol

The molar concentration of N2 is:

[N2] = n(N2) / V = 0.521 mol / 1.0 L = 0.521 M

For H2:

n(H2) = PV / RT = (2.0x10^7 Pa)(1.0 L) / (8.31 J/mol/K)(593 K) = 1.563 mol

The molar concentration of H2 is:

[H2] = n(H2) / V = 1.563 mol / 1.0 L = 1.563 M

For NH3:

n(NH3) = 1.5 mol

The molar concentration of NH3 is:

[NH3] = n(NH3) / V = 1.5 mol / 1.0 L = 1.5 M