Question

A vessel of volume 22.4 dm3 contains 2.0 mol H2 and 1.0 mol N2 at 273.15 K initially. All the H2 reacted with sufficient N2 to form NH3. Calculate the partial pressures and the total pressure of the final mixture.

Answer 1

the reaction is

3H₂ + N₂  $\rightarrow$ 2NH₃

According to reaction 3 mole of hydrogen react with 1 mole of nitrogen to form 1 mole of ammonia. then,

we have 2 mole of hydrogen hence, 2 mole of hydrogen  react with 0.67 mole of nitrogen to form 0.67 mole of ammonia and 0.33 mole of nitrogen remain unreacted.

final mixture contain 0.67 mole of ammonia and

0.33 mole of nitrogen.

total mole of gas in final mixture is 0.67 mole + 0.33 mole = 1 mole

According to ideal gas law 1 mole of gas at STP occupy volume 22.4 dm3, given tempreture is 273.5, volume 22.4 dm3 and gas is 1 mole thus, pressure is 1 atm.and partial pressure is

NH3 = 0.67 atm

N₂ = 0.33 atm

we can conform it by using ideal gas equation given as below

we know that ideal gas equation PV = nRT

P = nRT/V

= 1$\times$0.08205$\times$273.15/22.4 = 1