Question

For a first order reaction, rate constant is represented by $\log K\left( {{s}^{-1}} \right)=15-\dfrac{9.19\times {{10}^{3}}\left( K \right)}{T}$ , then the temperature at which its half life is 28 min (ln2 = 0.7, log3 = 0.48, log2 = 0.30)
(A) 500 K
(B) 919 K
(C) 1000 K
(D) 638 K

Answer 1

As we know that the given illustration is based on the first order reaction and the half life of the compound undergoing first order reaction. Thus, having basic ideas about the given parameters can help us solve the same.

Complete step by step solution:
Let us firstly see what we mean by first order reactions and the half life determination, then we will move forward towards the illustration;
First order reactions-
It is the reaction that depends on the concentration of only one reactant i.e. the overall rate of the reaction depends solely on the concentration of only one reactant.
This can be represented by,
$Rate=-\dfrac{d\left[ X \right]}{dt}=k\left[ X \right]$
Rearranging the above equation we get,
$\dfrac{d\left[ X \right]}{\left[ X \right]}=-kdt$
Now, integrating we get,
$\ln \left[ X \right]=\ln {{\left[ X \right]}_{0}}-kt$
By logarithmic theorems,
$\ln \left( \dfrac{{{\left[ X \right]}_{t}}}{{{\left[ X \right]}_{0}}} \right)=-kt$
We can also say that,
${{\left[ X \right]}_{t}}={{\left[ X \right]}_{0}}{{e}^{-kt}}$
We can say that there are many forms of the equation to represent the first order reactions. But the above given is most suitable hence,
The half life for the first order can be stated as;
The half life of any compound is the time to which the concentration of the same is reduced by half of its initial concentration i.e.
${{\left[ X \right]}_{\dfrac{1}{2}}}=\dfrac{1}{2}{{\left[ X \right]}_{0}}$
Thus,
$\dfrac{{{\left[ X \right]}_{\dfrac{1}{2}}}}{{{\left[ X \right]}_{0}}}=\dfrac{1}{2}={{e}^{-k{{t}_{\dfrac{1}{2}}}}}$
Solving the above equation we get,
${{t}_{\dfrac{1}{2}}}=\dfrac{\ln 2}{k}=\dfrac{0.693}{k}$
Now, using all the stated equations as above;
Illustration-
Given that,
Half life = 28 min = 1680 sec
Rate constant is represented as - $\log K\left( {{s}^{-1}} \right)=15-\dfrac{9.19\times {{10}^{3}}\left( K \right)}{T}$
Thus, solving we get,
$\begin{aligned} & {{t}_{{}^{1}/{}_{2}}}=\dfrac{0.693}{k} \\\ & k=\dfrac{0.693}{{{t}_{{}^{1}/{}_{2}}}} \\\ & k=\dfrac{0.693}{1680}=4.125\times {{10}^{-4}}{{\sec }^{-1}} \\\ \end{aligned}$
Now, as the given equation T can be stated as,
$\begin{aligned} & \log K\left( {{s}^{-1}} \right)=15-\dfrac{9.19\times {{10}^{3}}\left( K \right)}{T} \\\ & \log \left( 4.125\times {{10}^{-4}} \right)=-3.385=15-\dfrac{9.19\times {{10}^{3}}}{T} \\\ & T=500K \\\ \end{aligned}$

Therefore, we can say that option (A) is correct.

Note: Do note to use proper units as the given time is in minutes and the required rate constant value should be in the ${{s}^{-1}}$ . Thus, we need to have a sharp look towards the given data and generality to solve the equations in SI units.