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Question: The rate constant of a zero order reaction is \(0.2 mol dm^{-3} h^{-1}\). If the concentration of th...

The rate constant of a zero order reaction is 0.2moldm−3h−10.2 mol dm^{-3} h^{-1}. If the concentration of the reactant after 30 minutes is 0.05moldm−30.05 mol dm^{-3}. Then its initial concentration would be:
(A)0.15moldm−3(A) 0.15mold{m^{ - 3}}
(B)1.05moldm−3(B) 1.05mold{m^{ - 3}}
(C)0.25moldm−3(C) 0.25mold{m^{ - 3}}
(D)4.00moldm−3(D) 4.00mold{m^{ - 3}}

Explanation

Solution

Order of reaction is an experimentally determined quantity. It is defined as the sum of the power of concentration terms in the rate law expression. It can be either zero, first, second or third order. The order can also be fractional. In the question we are given zero order reaction. Zero order reactions are those in which the rate of reaction does not depend on or independent of the concentration terms.

Complete answer:
Consider a reaction,
2A+B→Product2A + B \to \text{Product}
Experimentally it was found that
Rate of reaction α[A]x\alpha {\left[ A \right]^x}
Rate of reactionα[B]y\alpha {\left[ B \right]^y}
Overall rate law expression will be, Rate=k[A]x[B]yRate = k{\left[ A \right]^x}{\left[ B \right]^y}
Where xx is the order of reaction w.r.t. AA and yy is the order of reaction w.r.t to BB
So, n=x+yn = x + y is the overall order of reaction
In case of zero order reaction,
(1)The unit of the rate of reaction is MoleL−1time−1Mole{L^{ - 1}}tim{e^{ - 1}}
(2)A→PA \to P suppose reactant AA are changing into product PP
In case of zero order reaction,
Rate=k[A]0−−−−(1)Rate = k{\left[ A \right]^0} - - - - (1)
It can also be written as,−d[A]dt=k[A]0=k\dfrac{{ - d\left[ A \right]}}{{dt}} = k{\left[ A \right]^0} = k
−d[A]=kdt- d\left[ A \right] = kdt
Integrating both sides within limits, ∫A0A−d[A]=∫0tkdt\int\limits_{{A_0}}^A { - d\left[ A \right]} = \int\limits_0^t {kdt}
[A0]−[A]=kt\left[ {{A_0}} \right] - \left[ A \right] = kt
k=[A0]−[A]t−−−−(2)k = \dfrac{{\left[ {{A_0}} \right] - \left[ A \right]}}{t} - - - - (2)
In the question we are given,
k=0.2moldm−3h−1k = 0.2mold{m^{ - 3}}{h^{ - 1}}
t=30min⁡=30÷60sec⁡=0.5ht = 30\min = 30 \div 60\sec = 0.5h
[A]=0.05moldm−3\left[ A \right] = 0.05mold{m^{ - 3}}, [A0]=?\left[ {{A_0}} \right] = ?
0.2=[A0]−[0.05]0.5=0.15moldm−30.2 = \dfrac{{\left[ {{A_0}} \right] - \left[ {0.05} \right]}}{{0.5}} = 0.15mold{m^{ - 3}}
So the correct answer is Option (A).

Note:
In the question we should be very careful of the units, always compare units in the question and that required in the solution to prevent any error. Some of the common examples of zero order reactions are photochemical reactions of Hydrogen and chlorine, decomposition of nitrous oxide over a hot palladium surface etc.