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For the gaseous reaction CH3CHO(g) -> CO(g) + CH4(g), the initial pressure is 80 mmHg. The total pressure after 20 minutes is 120 mmHg. Assuming first-order kinetics, find the rate constant.

1. 3.465 * 10^(-2) min^(-1)
2. 34.65 min^(-1)
3. 3.465 min^(-1)
4. 0.3465 min^(-1)

Correct answer: 3.465 * 10^(-2) min^(-1)

Solution

CH3CHO -> CO + CH4: 1 mole -> 2 moles. At time t, let x = pressure decrease of CH3CHO. Total pressure = (80-x) + x + x = 80+x. At t=20: 80+x=120 => x=40. Remaining P(CH3CHO) = 80-40 = 40 mmHg. k = (1/20)*ln(80/40) = (1/20)*ln2 = 0.693/20 = 0.03465 min^(-1) = 3.465*10^(-2) min^(-1).

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