Question

For a first order reaction rate constant is given as $\log K=14-\dfrac{1.2\times {{10}^{4}}}{T}$​ then what will be value of temperature if its half life period is $6.93\times {{10}^{-3}}min$?
A. $100K$
B. $1000K$
C. $720K$
D. $327K$

Answer 1

A first-order reaction can be defined as a chemical reaction in which the rate of reaction is linearly dependent on the concentration of only one reactant. In other words, a first-order reaction is a chemical reaction in which the rate of the reaction varies based on the changes in the concentration of only one of the reactants. Thus, the order of these reactions is equal to 1

Complete step by step answer:
Here we have given the first order reaction with a half-life period of $6.93\times {{10}^{-3}}min$. For the first order reaction, the equation for the rate constant is given below;
$K= \dfrac{0.693}{{{t}_{1/2}}}$
$K=$ Rate constant
${{t}_{1/2}}=$ Half-life of the reaction
Here, we have given that the half-life period is equal to $6.93\times {{10}^{-3}}min$. Substitute this value in the above equation.
$K=\dfrac{0.693}{6.93\times {{10}^{-3}}min}=100{{\min }^{-1}}$
We got the value of rate constant is equal to $100{{\min }^{-1}}$
Hence, we have to find the value of temperature from the equation ‘$\log K=14-\dfrac{1.2\times {{10}^{4}}}{T}$’
$\Rightarrow \log K=14-\dfrac{1.2\times {{10}^{4}}}{T}$
Here we have known the value of ‘$K$’ is equal to $100{{\min }^{-1}}$ ,put this value in above equation;
$\log (100{{\min }^{-1}})=14-\dfrac{1.2\times {{10}^{4}}}{T}$
$2=14-\dfrac{1.2\times {{10}^{4}}}{T}$
$\Rightarrow T=\dfrac{1.2\times {{10}^{4}}}{12}=1000K$
Here, we got the value of the temperature is equal to $1000K$

So, the correct answer is Option B.

Note: First-order reactions are very common. We have two examples of first-order reactions: the hydrolysis of aspirin and the reaction of t-butyl bromide with water to give t-butanol. Another reaction that exhibits apparent first-order kinetics is the hydrolysis of the anticancer drug cisplatin