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Question: The rate constant for a first order reaction is \(60{s^{ - 1}}\). How much time will it take to redu...

The rate constant for a first order reaction is 60s−160{s^{ - 1}}. How much time will it take to reduce the initial concentration of the reactant to its 1/10th value?

Explanation

Solution

The rate law of first order reaction is dependent on the time which means when the value of time changes, the rate constant value also changes. The time will be calculated by the formula of rate law. The final concentration will be 1/10th of the initial concentration.

Complete step by step answer:
The rate law is an equation which gives a relation between the rate of chemical reaction with the concentration of the reactant or the partial pressure of the reactant.
The reaction is said as a first order reaction when the reaction rate depends on the concentration of only one reactant. The unit of first order reaction is s−1{s^{ - 1}}.
Given,
The rate constant for first order reaction is 60s−160{s^{ - 1}}.
Let the initial concentration be aa.
Therefore, the final concentration (a-x) =110×a= \dfrac{1}{{10}} \times a
⇒\RightarrowThe final concentration (a-x) =a10= \dfrac{a}{{10}}
The rate constant is given by the formula as shown below.
K=2.303tlog⁡aa−xK = \dfrac{{2.303}}{t}\log \dfrac{a}{{a - x}}
Where,
K is the rate constant.
t is the time
aa is the initial concentration.
a−xa - x is the final concentration.
The formula for rate constant is rearranged to form the formula for time. Thus, the formula for calculating time is given as shown below.
t=2.303Klog⁡aa−xt = \dfrac{{2.303}}{K}\log \dfrac{a}{{a - x}}
To find the value of time, substitute the values in the above equation.
⇒t=2.30360log⁡aa10\Rightarrow t = \dfrac{{2.303}}{{60}}\log \dfrac{a}{{\dfrac{a}{{10}}}}
⇒t=2.30360log⁡10\Rightarrow t = \dfrac{{2.303}}{{60}}\log 10
⇒t=0.03838×log⁡10\Rightarrow t = 0.03838 \times \log 10
The value of log10 is 1, substitute the value of log 10
⇒t=0.03838×1\Rightarrow t = 0.03838 \times 1
⇒t=3.8×10−2\Rightarrow t = 3.8 \times {10^{ - 2}}s
Therefore, the time needed to reduce the initial concentration of the reactant to its 1/10th value is 3.8×10−23.8 \times {10^{ - 2}}.

Note:
For a first order reaction, the rate constant has only a time unit, it does not have a concentration unit which means that the value of rate constant k for the first order reaction is independent where the concentration is expressed.