Question

a student ran the following action in the laboratory at 662 k
2NH3 to N2 + 3H2
when he introduced an H3 at a pressure of 0.625 ATM into a 1-liter evacuated container he found the equilibrium partial pressure of H2 to be 0.923 ATM calculate the equilibrium constant KP he obtained for this reaction

Answer 1

Kp relates products and reactants pressure

for the balanced reaction

2NH3 <-> N2 + 3H2

we will have

Kp = P-N2 * (P-H2)^3 / (P-NH3)^2

Where P-X is the partial pressure of the component in equilibrium

so... we must calculate pressures in equilibrium

initally

P-NH3 = 0.625 atm

P-N2 = 0 atm

P-H2 = 0 atm

in equilibrium ( extent of reaction)

P-NH3 = 0.625 atm - 2x

P-N2 = 0 atm + x

P-H2 = 0 atm +3x

and we know that P-H2 = 0.923 atm

so

P-H2 = 0 atm +3x = 0.923 atm

solve for x

x = 0.923/3 = 0.30766

now, substitute in all x

P-NH3 = 0.625 atm - 2*0.30766 = 0.625  - 2*0.30766 = 0.00968 atm

P-N2 = 0 atm + x = 0.30766 atm

P-H2 = 0 atm +3x = 3*0.30766 = 0.92298 atm

now that we have all equilibrium pressures:

Kp = P-N2 * (P-H2)^3 / (P-NH3)^2

Kp = 0.30766 * (0.92298 )^3 / (0.00968 )^2

Kp = 2581.64

which means, the reaction favours strongly the decomposition of NH3 at these conditions