Question

A first-order reaction has a rate constant of 0.33 min-1. It takes __________ min for the reactant concentration to decrease from 0.13 M to 0.088 M

Answer 1

Answer on Question #69833 - Chemistry - Inorganic Chemistry

Task:

A first-order reaction has a rate constant of $0.33\ \mathrm{min}^{-1}$. It takes ___ min for the reactant concentration to decrease from 0.13 M to 0.088 M

Solution:

For the reaction $\mathrm{aA} + \mathrm{bB} = \mathrm{cC} + \mathrm{dD}$

$$\text{rate} = -\frac{1}{a} \frac{\Delta[A]}{\Delta t} = -\frac{1}{b} \frac{\Delta[B]}{\Delta t} = +\frac{1}{c} \frac{\Delta[C]}{\Delta t} + \frac{1}{d} \frac{\Delta[D]}{\Delta t}$$ $$K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}$$

$1^{\text{st}}$ order reactions:

$$\ln[\mathrm{A}]_t = -\mathrm{k}t + \ln[\mathrm{A}]_0$$ $$t_{1/2} = 0.693/\mathrm{k}$$

Then,

$$k = 0.33\ \mathrm{min}^{-1}; \ln[A]_0 = 0.13\ \mathrm{M}; \ln[A] = 0.088\ \mathrm{M};$$ $$\ln[A]_0 - \ln[A] = \mathrm{k}t;$$ $$\ln[0.13] - \ln[0.088] = 0.33*\mathrm{t};$$ $$-2.04 - (-2.43) = 0.33*\mathrm{t};$$ $$t = 1.18\ \mathrm{min}.$$

**Answer:** 1.18 min.

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