Question

For the reaction, a → b, the rate constant is 0.0208 m-1 sec-1. How long would it take for [a] to decrease from 0.100 to 0.0450 m?

Answer 1

First, assume the order of the given reaction is n, then the rate of reaction i.e. $\frac{dx}{dt}=k\times[A]^{n}$

where, dx is change in concentration of A in small time interval dt and k is rate constant.

According to units of rate constant, the reaction is of second order.

$\frac{1}{[A]_{t}}-\frac{1}{[A]_{o}} = kt$   (second order formula)

Put the values,

$\frac{1}{0.04590 m}-\frac{1}{0.100 m} =0.0208 m^{-1}s^{-1} \times t$  

 $22.23 m -10 m =0.0208 m^{-1}s^{-1} \times t$

$\frac{12.23 m}{0.0208 m^{-1}s^{-1}} = t$

t= 587.9 s

Hence, time taken is 587.9 s

Answer 2

The rate constant is the constant value of proportionality that defines the rate of reaction with the concentration. It will take 587.9 seconds for a to decrease.

What is a rate constant?

The rate constant is the relation of the rate of the chemical reaction with that of the concentration of the reactants used in the reaction of the product formation.

When the order of the reaction is "n" then the rate of reaction is given as,

$\rm \dfrac{dx}{dt} = k\times [A]^{n}$

Here, dx is concentration change, A is the concentration and dt and k is rate constant.

The reaction is of second-order as the unit of the rate constant is per meter per second.

The formula for the second-order reaction will be:

$\rm \dfrac{1}{[A]_{t}} - \rm \dfrac{1}{[A]_{o}} = \rm kt$

Substituting values in the equation:

$\begin{aligned}\dfrac{1}{0.04590} - \dfrac{1}{0.100} &= 0.0208 \;\rm m^{-1}s^{-1} \times t\\\\& = 22.23 - 10 = 0.0208 \times \rm t\\\\&\rm t = 587.9 \;\rm s\end{aligned}$

Therefore, the time taken by a is 587.9 seconds.

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