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Question: In a zero-order reaction, \[20\% \] of the reaction is complete in \[10\;s\]. How much time will it ...

In a zero-order reaction, 20%20\% of the reaction is complete in 10  s10\;s. How much time will it take to complete 50%50\% of the reaction?

  1. 20  s20\;s
  2. 25  s25\;s
  3. 30  s30\;s
  4. 40  s40\;s
Explanation

Solution

Zero-order reaction: It is a reaction in which the rate of formation of a product does not depend on the concentration of any of the reactants and the rate constant remains fixed throughout the reaction process. In zero-order reactions, if the concentration versus time graph is plotted then a straight-line curve is obtained.

Complete answer:
Consider a zero-order reaction A→BA \to B,
The differential rate law for the zero-order reaction can be written as follows:
−d[A]dt=k[A]0- \dfrac{{d[A]}}{{dt}} = k{[A]^0}
On solving above equation, the integrated rate law for the zero-order reaction obtained is as follows:
d[A][A]0=−kdt\dfrac{{d[A]}}{{{{[A]}^0}}} = - kdt
⇒∫[A]o[A]td[A]=−k∫0t dt\Rightarrow \int\limits_{{{[A]}_o}}^{{{[A]}_t}} {d[A]} = - k\int\limits_0^t {\,dt}
⇒[A]t−[A]o=−kt\Rightarrow {[A]_t} - {[A]_o} = - kt
⇒[A]o−[A]t=kt     −(i)\Rightarrow {[A]_o} - {[A]_t} = kt\,\;\; - (i)
Where, [A]o{[A]_o} is the initial amount of reactant, [A]t{[A]_t} is the amount of reactant left, kk is the rate constant and tt is the time taken to complete the reaction.
As per question, time taken to complete 20%20\% of the reaction =10  s = 10\;s.
That means after ten seconds, the amount of reactant left in the reaction =1−20100[A]o = 1 - \dfrac{{20}}{{100}}{[A]_o}
⇒[A]t=1−0.2[A]o\Rightarrow {[A]_t} = 1 - 0.2{[A]_o}
⇒[A]t=0.8[A]o\Rightarrow {[A]_t} = 0.8{[A]_o}
Substituting values in equation (i):
[A]o−0.8[A]o=k×10{[A]_o} - 0.8{[A]_o} = k \times 10
⇒k=0.2[A]o10\Rightarrow k = \dfrac{{0.2{{[A]}_o}}}{{10}}
⇒k=0.02[A]o  −(ii)\Rightarrow k = 0.02{[A]_o}\,\, - (ii)
Now, it is given that 50%50\% of the reaction is completed i.e., the amount of reactant left will be as follows:
[A]t=1−0.5[A]o{[A]_t} = 1 - 0.5{[A]_o}
⇒[A]t=0.5[A]o\Rightarrow {[A]_t} = 0.5{[A]_o}
Substituting values in equation (i):
[A]o−0.5[A]o=kt{[A]_o} - 0.5{[A]_o} = kt
⇒0.5[A]o=kt\Rightarrow 0.5{[A]_o} = kt
Substituting value of kkfrom equation (ii):
⇒0.5[A]o=0.02[A]o×t\Rightarrow 0.5{[A]_o} = 0.02{[A]_o} \times t
⇒t=0.50.02\Rightarrow t = \dfrac{{0.5}}{{0.02}}
⇒t=25 s\Rightarrow t = 25\,s
Therefore, the time required to complete 50%50\% of the zero-order reaction is 25  s25\;s.
Hence, option (2) is the correct answer.

Note:
Do not get confused with the value of [A]t{[A]_t}. It is the amount or concentration of reactant left after the reaction, not the concentration of reactant consumed during the reaction. So, substitute values accordingly. Also, the unit of rate constant for zero-order reaction is molL−1s−1mol{L^{ - 1}}{s^{ - 1}}.