Question

If $50\% $ of a reaction occurs in ${\text{100}}$ second and $75\% $ of the reaction occurs in ${\text{200}}$ second, the order of this reaction is:
A) ${\text{1}}$
B) ${\text{0}}$
C) ${\text{2}}$
D) ${\text{3}}$

Answer 1

Hint: The order of reaction of total reaction has been asked and one can find out that by doing the separate calculation of the order of reaction for both reactions. The calculation results can be related at last and order can be determined. If the calculations are the same then the order will be one and for if the calculation value gets doubled then the reaction will be the second-order reaction.

Complete step by step answer:
1) First of all we will learn what order of the reaction is, where the Order of reaction can be said as the relationship between the rate of a chemical reaction and the concentration of the chemical entities which are taking part in the reaction.
2) calculate the order of reaction by using the formula,
$k = \dfrac{{2 \cdot 303}}{t}\log \dfrac{{{{\left[ A \right]}_0}}}{{{{\left[ A \right]}_t}}}$
Where, ${\text{k}}$= rate of reaction, ${\text{t}}$= time for that reaction, ${\left[ A \right]_0}$= initial concentration of reaction, ${\left[ A \right]_t}$= final concentration of reaction.
3) Now let’s calculate the rate of reaction for the first part where $50\% $ of a reaction occurs in ${\text{100}}$ second as follows,
$k = \dfrac{{2 \cdot 303}}{t}\log \dfrac{{{{\left[ A \right]}_0}}}{{{{\left[ A \right]}_t}}}$
$k = \dfrac{{2 \cdot 303}}{{100}}\log \dfrac{{100}}{{50}}$
We have taken initial concentration as ${\text{100}}$ because initially the concentration amount is full and always taken as ${\text{100}}$ and as the $50\% $ of a reaction has happened the final concentration will drop to ${\text{50}}$.
$k = \dfrac{{2 \cdot 303}}{{100}}\log \left( 2 \right)$
$k = \dfrac{{2 \cdot 303}}{{100}} \times 0 \cdot 3010$
$k = 0 \cdot 00693$
Therefore, the rate of reaction for the first part where $50\% $ of a reaction occurs in ${\text{100}}$ second is $0 \cdot 00693$
4) Now let’s calculate the rate of reaction for the second part where $75\% $ of a reaction occurs in ${\text{200}}$ second as follows,
$k = \dfrac{{2 \cdot 303}}{t}\log \dfrac{{{{\left[ A \right]}_0}}}{{{{\left[ A \right]}_t}}}$
$k = \dfrac{{2 \cdot 303}}{{200}}\log \dfrac{{100}}{{25}}$
We have taken initial concentration as ${\text{100}}$ because initially the concentration amount is full and always taken as ${\text{100}}$ and as the $75\% $ of a reaction has happened the final concentration will drop to ${\text{25}}$.
$k = \dfrac{{2 \cdot 303}}{{200}}\log \left( 4 \right)$
$k = \dfrac{{2 \cdot 303}}{{200}} \times 0 \cdot 6020$
$k = 0 \cdot 00693$
Therefore, the rate of reaction for the second part where $75\% $ of a reaction occurs in ${\text{200}}$ second is $0 \cdot 00693$
5) Now that both reactions give the rate of reaction $0 \cdot 00693$ and the value of ${\text{k}}$ is constant the reaction is first-order reaction

So, the correct answer is Option A .

Note:
The order of a reaction is the rate that is dependent on the concentration of chemical species and in the first-order reaction the rate will be dependent on the only concentration of one reactant. Due to the dependence of rate on only one chemical species the rate of reaction will be the same even if we make changes to the time of reaction or the percentage consumed in that reaction.