Question

For the following reaction;
$\begin{array}{l}2X + Y\overset{k}{\rightarrow}P\end{array}$

The rate of reaction is
$\begin{array}{l}\frac{d[P]}{dt} = k[X]\end{array}$
Two moles of X are mixed with one mole of Y to make 1.0 L of solution. At 50 s, 0.5 mole of Y is left in the reaction mixture. The correct statement(s) about the reaction is(are). (Use: ln 2 = 0.693)

• A. The rate constant, k, of the reaction is 13.86 × 10-4 s-1.
• B. Half-life of X is 50 s.
• C. At 50 s, – d[X] / dt = 13.86 × 10-3 mol L-1 s-1.
• D. At 100 s, d[Y] / dt = 3.46 × 10-3 mol L-1 s-1.

Answer 1

Answer: (B), (C), (D)

Half-life of X is 50 s.

Answer 2

rate =$\begin{array}{l}\frac{d[P]}{dt} = k[X]\end{array}$
$2X + Y → P$
2 mole 1 mole
1 mole 0.5 mole 0.5 mole
$- \frac{{d[X]}}{{dt}} = k_1[X] = 2k[X] \Rightarrow 2k = k_1$
$-\frac{d[Y]}{dt} = k_2[X] = 2k[X] \Rightarrow k_2 = k$
$2k = \frac{1}{50} \ln 2$
$K = \frac{1}{100} \ln 2 = 0.693 \times 10^{-2} \times \text{s}^{-1} = 50 \, \text{sec}^{-1}$
At 50 sec,
$\frac{d[X]}{dt} = 2k[X] = 2 \times \frac{0.693}{100} \times 1$
= $13.86 \times 10^{-3} \, \text{mol} \, \text{L}^{-1} \, \text{s}^{-1}$
At 100 sec
$-\frac{d[Y]}{dt} = k_2[X] = k[X] \times \frac{0.693}{100} \times \frac{1}{2}$
(Concentration of X after 2 half-lives = ½ M)
=$$3.46 \times 10^{-3} \, \text{mol} \, \text{L}^{-1} \, \text{s}^{-1}