Question

A first-order reaction, ${\text{A}} \to {\text{B}}$ , requires activation energy of $70{\text{ kJmo}}{{\text{l}}^{ - 1}}$. When a $20\%$ solution of A was kept at $25^\circ C$for 20 min, $25\%$ decomposition took place. What will be the percentage decomposition at the same time in a $30\%$ solution maintained at $40^\circ C$? Assume that activation energy remains constant in this range of temperature.
A.Decomposition $67.21\%$
B.Decomposition $33.5\%$
C.Decomposition $50\%$
D.None of these

Answer 1

To answer this question, you should recall the concept of first-order reactions. The first-order decomposition can be defined as the chemical reaction involving only one chemical species, in which the rate of decrease of the concentration of the reactant is directly proportional to its concentration. Radioactivity is an example of the first-order decomposition.

Formula used: ${\text{K}} = \dfrac{{2.303}}{{\text{t}}}{\text{log}}\dfrac{{\text{a}}}{{{\text{a}} - {\text{X}}}}$where ${\text{K}}$is the rate coefficient, ${\text{t}}$is time, ${\text{a}}$is initial concentration and ${\text{a}} - {\text{X}}$is final concentration

Complete step by step solution:
We are given the change, ${\text{A}} \to {\text{B}}$ in which $20\%$ solution of A decomposes $25\%$ in 20 minutes at ${25^o}C$. Let the amount of solute undergoing decay be ${\text{a}}$=20 units.
∴ Amount of solute left ${\text{a}} - {\text{X}}$ after 20 minutes will be equal to $75\%$of initial amount.
$\Rightarrow 20 \times \dfrac{{75}}{{100}} =$15.
Now calculating the rate of reaction using equation (i)
∴${{\text{K}}_{{\text{25}}}}{\text{ }} = {\text{ }}\dfrac{{2.303}}{{\text{t}}}\log \dfrac{{\text{a}}}{{{\text{a - X}}}}{\text{ }} = {\text{ }}\dfrac{{2.303}}{{20}}{\text{log}}\dfrac{{20}}{{15}}$
$\Rightarrow 1.44 \times {10^{ - 2}}{\text{ minut}}{{\text{e}}^{ - 1}} = 0.0144{\text{ minut}}{{\text{e}}^{{\text{ - 1}}}}$.

Now we will calculate the rate coefficient for the second condition i.e. $40^\circ C$.
Here ${T_1} = 25^\circ C = 298K$and ${T_2} = {40^o}C = 313K$
$2.303{\text{log}}\dfrac{{{{\text{K}}_{{\text{40}}}}}}{{{{\text{K}}_{{\text{25}}}}}}{\text{ = }}\dfrac{{{{\text{E}}_{\text{a}}}}}{{\text{R}}}\,\,\left[ {\dfrac{{{{\text{T}}_{\text{2}}}{\text{ - }}{{\text{T}}_{\text{1}}}}}{{{{\text{T}}_{\text{1}}}{ \times }{{\text{T}}_{\text{2}}}}}} \right]$
Substituting the values:
$\Rightarrow 2.303{\text{log}}\dfrac{{{{\text{K}}_{{\text{40}}}}}}{{0.0144}}{\text{ }} = {\text{ }}\dfrac{{70 \times {{10}^3}}}{{8.314}}\left[ {\dfrac{{313 - 298}}{{313 \times 298}}} \right]$
Solving this, we get:
∴${{\text{K}}_{{\text{40}}}} = 5.58 \times {10^{ - 2}}{\text{ minut}}{{\text{e}}^{{\text{ - 1}}}}$

Now, applying equation (i) for the second given condition, let the left out concentration be $m$after 20 minutes.
$\Rightarrow {{\text{K}}_{{\text{40}}}} = \dfrac{{2.303}}{{\text{t}}}{\text{log}}\dfrac{{\text{a}}}{{\left( {{\text{a}} - {\text{X}}} \right)}}$​
$\Rightarrow 5.58 \times {10^{ - 2}} = \dfrac{{2.303}}{{20}}{\text{log}}\dfrac{{30}}{{\text{m}}}$​
Solving this:
$\Rightarrow m = 9.83$

$\therefore$Finally, the percentage of decomposition can be calculated by dividing the final concentration with initial concentration:
$\dfrac{{{\text{a}} - {\text{m}}}}{{\text{a}}} \times 100$
$\Rightarrow \dfrac{{30 - 9.83}}{{30}} \times 100 = 67.2\%$

Therefore, we can conclude that the correct answer to this question is option A.

Note: You should know about the difference in molecularity and order of a reaction. The molecularity can be defined as the number of atoms, molecules, or ions which must undergo a collision with each other in a short time interval for the chemical reaction to proceed. Unlike the order of a reaction, it applies to simple reactions only.