Question

In a zero-order reaction, \[47.5\% \] of the reactant remains at the end of 2.5 hours. The amount of reactant consumed in one hour is
A. \[10.5\% \]
B. \[32.0\% \]
C. \[52.6\% \]
D. \[21.0\% \]

Answer 1

Hint: Since, it is given in the question that at the end of 2.5 hours, \[47.5\% \] of the reactant remains in a zero-order reaction. So, we will take the initial amount, ${A^0}$ to be 100. Therefore, the right amount, ‘A’ will be 47.5. So, substituting the values in the equation \[A = {A^0} - kt\], we will get calculate k, and after getting the value of k, we will put it in the same equation in order to find the amount of reactant consumed in one hour i.e. \[t = 1\] hour.

Complete step by step answer:
Given in the question is,
In a zero-order reaction, \[47.5\% \] of the reactant remains
i.e. left amount \[A = 47.5\% \]
Initial amount \[{A^0}\; = 100\]
and the time taken hours
From the zero-order reaction we know,
 \[A = {A^0} - kt\]
After substituting the values in the above equation, we get
 \[47.5 = 100 - k \times 2.5\]
On solving we get,
 \[k = 21\]
Similarly, to find the amount of reactant consumed in one hour,
Initial amount \[{A^0}\; = 100\]
and the time taken hours
 \[A - {A^0} = ?\]
Putting values in the equation \[A = {A^0} - kt\], we get
 \[A - {A^0} = 21 \times 1 = 21\% \]
So, the amount of reactant consumed in one hour is \[21\% \].

Therefore, the correct answer is option (D).

Note: Zero - order reaction is a chemical reaction where the rate does not vary with the increase or decrease in the concentration of the reactants. Therefore, the rate of these reactions is always equal to the rate constant of the specific reactions since the rate of these reactions is proportional to the zeroth power of reactants concentration.