Question

2NH3 —> N2 (g) + 3H2 (g)

If the total pressure of the gas at the end if the reaction is 759 mmHg, calculate the partial pressures in atm of nitrogen(N2) and hydrogen (H2)

Please explain so I can understand how the answer was derived

Answer 1

The given reaction is--

2NH₃ ------> N₂(g) + 3H₂(g)

Here we can see that there is 1 mol of N₂ and 3 mol of H₂present.

Total number of moles = 1 + 3 = 4

Given, P_total = 759 mm Hg

Now, mole of N₂ present / total number of moles = partial pressure of N₂ / P_total

=> 1 / 4 = partial pressure of N₂ / 759 mm Hg

=> partial pressure of N₂ = 759 mm Hg / 4

=> partial pressure of N₂ = 189.75 mm Hg = 189.75 mm Hg * ( 1atm /760 mm Hg) =0.249 atm

Again,

mole of H₂ present / total number of moles = partial pressure of H₂ / P_total

=> 3/ 4 = partial pressure of H₂ / 759 mm Hg

=> partial pressure of H₂ =( 759 mm Hg x 3) / 4

=> partial pressure of H₂ = (2277 / 4) mm Hg

=> partial pressure of H₂ = 569.25 mm Hg = 569.25 mm Hg *(1 atm /760 mm Hg) =0.749 atm