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A reaction follows the rate law d[P]/dt = k[X]. Two moles of X and one mole of Y are dissolved to make 1.0 L of solution. After 50 s, 0.5 mole of Y remains in the mixture. (Use ln 2 = 0.693) Which of the following statements about the reaction are correct? (More than one may be correct.)

1. The rate constant k of the reaction is 13.86 * 10⁻⁴ s⁻¹.
2. The half-life of X is 50 s.
3. At t = 50 s, the rate -d[X]/dt = 13.86 * 10⁻³ mol/L/s.
4. At t = 100 s, the rate -d[Y]/dt = 3.46 * 10⁻³ mol/L/s.

Correct answer: At t = 100 s, the rate -d[Y]/dt = 3.46 * 10⁻³ mol/L/s.

Solution

The reaction is first order in X. At 50 s, [X] drops from 2 M to 1 M (half-life = 50 s). k = ln2/t(1/2) = 0.693/50 = 13.86*10⁻³ s⁻¹. Option A gives k = 13.86*10⁻⁴ which is 10x too small. At 50 s, -d[X]/dt = k[X] = 13.86*10⁻³ * 1.0 = 13.86*10⁻³ mol/L/s (option C correct). At 100 s, [X] = 0.5 M, -d[X]/dt = 13.86*10⁻³*0.5 = 6.93*10⁻³. Since 2X -> Y (2:1 stoichiometry), -d[Y]/dt = (1/2)*(-d[X]/dt) = 3.46*10⁻³ mol/L/s (option D correct).

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