Question

Nitrogen and hydrogen react in the Haber process to form ammonia. All substances are in the gas phase. If 0.504 atm of nitrogen and 0.637 atm of hydrogen react, what is the partial pressure of ammonia (in mmHg) when this reaction goes 74.1 complete. The volume and temperature are constant.

Answer 1

N2+3H2----2NH3

Since, volume and temperature are constant, the partial pressure can be considered equivalent to number of moles and treated as such.

The reaction requires 3 times the number of moles of nitrogen than the moles of hydrogen to completely exhaust the reactants. Since, the partial pressure of hydrogen is less than 3 times of nitrogen partial pressure, N2 acts as the limiting agent. Therefore,

No. of moles of NH3 formed= 2 X (No. of moles of N2 used)

This is equivalent to,

Partial pressure of NH3= 2 * (Decrease in N2 partial pressure)= 2 * 0.504* 74.1/100=0.746928 atm =0.746928*760 mmHg= 567.67 mmHg