2NH3 —> N2 (g) + 3H2 (g) If the total pressure of the gas at the end if the reac

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

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


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

Explanation / Answer

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, Ptotal = 759 mm Hg

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

=> 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 / Ptotal

=> 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

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