Question

# #5.133 A mixture of NH3(g) and N2H4(g) is placed in a sealed container at 320 K...

#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.)

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

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