Question

A reaction is of first order. After 100 minutes 75 g of the reactant A are decomposed when 100 g are taken initially. Calculate the time required when 150 g of the reactant A are decomposed, the initial weight taken is 200 g:

A.100 minutes
B.200 minutes
C.150 minutes
D.175 minutes

Answer 1

Hint:A chemical reaction whose rate is dependent on concentration of only a reactant is termed as first order reaction. Here, we have to use the expression of integrated rate equation of first order reaction, that is, $k = \dfrac{{2.303}}{t}\log \dfrac{{{{\left[ A \right]}_0}}}{{\left[ A \right]}}$.

Complete step by step answer:

We know that, first order rate expression is,

$k = \dfrac{{2.303}}{t}\log \dfrac{{{{\left[ A \right]}_0}}}{{\left[ A \right]}}$ …… (1)

Where, k is the rate constant, t is time, ${\left[ A \right]_0}$ is initial concentration of the reactant and $\left[ A \right]$ is final concentration of the reactant.


Given that the reaction is a first order reaction. In the first case, decomposition of 75 g of the reactant A takes place when 100 g are taken initially in 100 minutes.

So, $t = 100\,\,\min $ , ${\left[ A \right]_0} = 100\,{\rm{g}}$ and $\left[ A \right] = 75\,{\rm{g}}$


Now, we have to put all the values in equation (1).

$k = \dfrac{{2.303}}{{100}}\log \dfrac{{100}}{{75}}$ …… (2)

In the second case, decomposition of 150 g of the reactant A takes place when 200 g are taken initially in t minutes. So, using equation (1),

$k = \dfrac{{2.303}}{t}\log \dfrac{{200}}{{150}}$ …… (3)

Now, we have to equate equation (2) and (3) as the rate constant is equal.

$ \Rightarrow \dfrac{{2.303}}{{100}}\log \dfrac{{100}}{{75}} = \dfrac{{2.303}}{t}\log \dfrac{{200}}{{150}}$

$ \Rightarrow \dfrac{1}{{100}}\log \dfrac{{100}}{{75}} = \dfrac{1}{t}\log \dfrac{{200}}{{150}}$

Now, we have to use the properties of logarithm.

$ \Rightarrow \dfrac{1}{{100}}\left( {\log 100 - \log 75} \right) = \dfrac{1}{t}\left( {\log 200 - \log 150} \right)$

$ \Rightarrow \dfrac{1}{{100}}\left( {2 - 1.88} \right) = \dfrac{1}{t}\left( {2.30 - 2.18} \right)$

$ \Rightarrow \dfrac{1}{{100}} \times 0.12 = \dfrac{1}{t} \times 0.12$

$ \Rightarrow t = 100\,\min $

Therefore, in 100 minutes decomposition of 150 g of reactant A takes place if 200 g of A is taken initially.

Hence, the correct answer is option A.

Note:
Always remember that, considering the rate law expression, summation of powers of the concentration of reactants gives the order of reaction. Order of reaction can be 0,1,2,3 or even a fraction. Zero order indicates that rate of reaction is independent of reactant concentration.