Question

#5.133

A mixture of NH3(g) and N2H4(g) is placed in a sealed container at 320 K . The total pressure is 0.53 bar . The container is heated to 1200 K at which time both substances decompose completely according to the equations 2NH3(g)→N2(g)+3H2(g) ; N2H4(g)→N2(g)+2H2(g) . After decomposition is complete the total pressure at 1200 K is found to be 4.5 bar.

Find the percent (by volume) of N2H4(g) in the original mixture. (Assume two significant figures for the temperature.)

Answer 1

0.51 atm at 320K, would be (0.51 atm) x (1200 / 320) =
Let z be the partial pressure (in atm) of N2H4 in the original mixture.
Then 0.51-z is the partial pressure of NH3 in the original mixture.
N2H4 → N2 + 2 H2
So z atm of H2N4 would produce 3z of products.
2 NH3 → N2 + 3 H2 [That is: 1 mol of NH3 produces 2 mol of products.]
So 0.51-z of NH3 would produce 1.02 - 2z of products.
(4.5 atm) x (320 / 1200) = 1.2 atm at 320K
Add the two expressions for partial pressure of products and set it equal to the total pressure of products at the same temperature:
3z + (1.02 - 2z) = 1.2 atm
Solve for z algebraically:
z = 0.18 atm originally
(0.18 / 0.51) = 0.35 = 35% N2H4 by partial pressure, and so 35% N2H4 by volume

Hope this helps