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For the first-order gas-phase reaction A(g) -> 2B(g) + C(g) at constant volume and 300 K, the total pressures at t = 0 and at time t are P0 and Pt respectively (initially only A is present). Let t(1/3) be the time for the partial pressure of A to fall to one-third of its initial value. Which statement is correct?

1. A plot of ln(P0 - Pt) versus time is a straight line with negative slope, because P0 - Pt is proportional to the amount of A reacted.
2. t(1/3) increases linearly with the initial concentration [A]0.
3. The rate constant increases with the initial concentration [A]0.
4. A plot of ln(P0 - Pt) versus time is a straight line with positive slope.

Correct answer: A plot of ln(P0 - Pt) versus time is a straight line with negative slope, because P0 - Pt is proportional to the amount of A reacted.

Solution

If A decreases by p, total pressure rises by 2p, so p = (Pt - P0)/2 and partial pressure of A = P0 - (Pt - P0)/2; the quantity P0 - Pt tracks A reacted, and since A is first order, ln of the remaining-A term falls linearly with time (negative slope), while t(1/3) and k are independent of [A]0.

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