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Question: In a zero-order reaction, \[47.5\% \] of the reactant remains at the end of 2.5 hours. The amount of...

In a zero-order reaction, 47.5%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%10.5\%
B. 32.0%32.0\%
C. 52.6%52.6\%
D. 21.0%21.0\%

Explanation

Solution

Since, it is given in the question that at the end of 2.5 hours, 47.5%47.5\% of the reactant remains in a zero-order reaction. So, we will take the initial amount, A0{A^0} to be 100. Therefore, the right amount, ‘A’ will be 47.5. So, substituting the values in the equation A=A0ktA = {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=1t = 1 hour.

Complete step by step answer:
Given in the question is,
In a zero-order reaction, 47.5%47.5\% of the reactant remains
i.e. left amount A=47.5%A = 47.5\%
Initial amount A0  =100{A^0}\; = 100
and the time taken hours
From the zero-order reaction we know,
A=A0ktA = {A^0} - kt
After substituting the values in the above equation, we get
47.5=100k×2.547.5 = 100 - k \times 2.5
On solving we get,
k=21k = 21
Similarly, to find the amount of reactant consumed in one hour,
Initial amount A0  =100{A^0}\; = 100
and the time taken hours
AA0=?A - {A^0} = ?
Putting values in the equation A=A0ktA = {A^0} - kt, we get
AA0=21×1=21%A - {A^0} = 21 \times 1 = 21\%
So, the amount of reactant consumed in one hour is 21%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.