Question

For a first order reaction, show that the time required for 99% completion is twice the time required for the completion of 90% reaction.

Answer 1

Hint: The rate constant in both the cases will be the same. To find the rate constant, the following relation is used:
$k=\dfrac{2.303}{t}\log \dfrac{a}{a-x}$….where, ‘a’ is the initial concentration
‘a-x’ is the concentration left
‘k’ is the rate constant.
‘t’ is the time required
We will calculate ‘k’ in both the cases and then equate them to get the desired result.

Complete step-by-step answer:
Let’s look at the derivation
To solve this question we will use the rate constant formula for first order reaction. In first order reaction the rate is dependent on the concentration of only one of the reactants.
Let ‘a’ be the initial concentration and ‘a-x’ be the amount of substance left in the reacting vessel.
Now we will take case1 for the completion of 99% of the reaction.
We will assume the initial concentration of the sample to be 100. This will make the conversation easy. Then we will subtract 1 from the initial concentration to get the final concentration.

Let a=100
then,a-x=100-99=1
$k=\dfrac{2.303}{{{t}_{99}}}\log \dfrac{a}{a-x}$
On putting the given values in the above equation, we get,
${{t}_{99}}=\dfrac{2.303}{k}\log \dfrac{100}{1}$
${{t}_{99}}=\dfrac{2.303}{k}\times 2$……..where $\log 100=2$
Hence, time required in this case is
${{t}_{99}}=\dfrac{2.303}{k}\times 2$……..equation 1
Now, we will consider the case for the completion of 90% of the reaction.
Let ‘a’ be the initial concentration and ‘a-x’ be the amount of substance left in the reacting vessel.
We will assume the initial concentration of the sample to be 100. This will make the conversation easy. Then we will subtract 10 from the initial concentration to get the final concentration.

Let a=100
then,a-x=100-90=10
$k=\dfrac{2.303}{{{t}_{90}}}\log \dfrac{a}{a-x}$
On putting the given values in the above equation, we get,
${{t}_{90}}=\dfrac{2.303}{k}\log \dfrac{100}{10}$……where $\log 10=1$
${{t}_{90}}=\dfrac{2.303}{k}\times 1$
Hence, time required in this case is
${{t}_{90}}=\dfrac{2.303}{k}\times 1$……equation 2
Now, from equation 1 and equation 2, we get,
${{t}_{99}}=2\times {{t}_{90}}$
Which is the required result.

Hence, the answer to the given question is ${{t}_{99}}=2\times {{t}_{90}}$

Note: You should transform the equations properly. The units of the rate constant vary with the order of the reaction. For first order reaction the unit of the rate constant is time inverse (${{\sec }^{-1}}$).