A mixture of H2 and O2 in 2:1 mole ratio is used to prepare water vapour by the reaction:
2H2(g) + O2(g) → 2H2O(g)
The total pressure of gases in the container is 4.5 atm at 57°C, before the reaction. The final total pressure of gases (in atm) at 127°C after reaction assuming 80% yield of water vapour is

Question

A mixture of H2 and O2 in 2:1 mole ratio is used to prepare water vapour by the reaction:

2H2(g) + O2(g) → 2H2O(g)

The total pressure of gases in the container is 4.5 atm at 57°C, before the reaction. The final total pressure of gases (in atm) at 127°C after reaction assuming 80% yield of water vapour is

TestHub · Accepted Answer

Before reaction (330 K): P(H2) = (2/3)(4.5) = 3.0 atm, P(O2) = 1.5 atm.

80% yield: H2 consumed = 0.8 × 3.0 = 2.4 atm-equivalent, O2 consumed = 1.2, H2O formed = 2.4.

Remaining at 330 K: H2 = 0.6, O2 = 0.3, H2O = 2.4; total = 3.3 atm.

Scale to 400 K: P_final = 3.3 × 400/330 = 4.0 atm.

Chemistry - Ideal Gas Question with Solution | TestHub

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