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Question: Reaction \(A + \,B \to C + D\) follows following rate law :\( = k{\left[ A \right]^{\ frac{1}{2}}}{\...

Reaction A+ B→C+DA + \,B \to C + D follows following rate law :=k[A] frac12[B]12 = k{\left[ A \right]^{\ frac{1}{2}}}{\left[ B \right]^{\frac{1}{2}}}starting with initial conc. of 1 mole1\,moleof A and B each ,what is the time taken for concentration of A to become 0.25 mole0.25\,mole?(Given:K=2.31 × 10−3 sec⁡−1K = 2.31\, \times \,{10^{ - 3}}\,{\sec ^{ - 1}})
A.300  sec⁡300\,\,\sec
B.600  sec⁡600\,\,\sec
C.900 sec⁡900\,\sec
D.None of these

Explanation

Solution

We know that in a chemical reaction, kinetics is always studied in terms of order. We may define order of a reaction as the sum of all the powers which are present on the concentration term in the rate law expression.

Complete Step by step answer: A reaction is said to be zero order, if the sum of all powers raised to the concentration is zero and if a reaction is said to be first order if the sum of all powers raised to the concentration is unity. If the sum of all the powers raised to concentration is two, then the reaction is said to be second order. If the sum of all the powers raised to concentration is three, then the reaction is said to be third order.
Given that; reaction A+ B→C+DA + \,B \to C + D takes place.
Rate law=k[A]12[B]12 = k{\left[ A \right]^{\frac{1}{2}}}{\left[ B \right]^{\frac{1}{2}}}
initial conc. of A is1 M1\,M.
initial conc. of B is 1 M1\,M.
rate constant for the reaction,K=2.31 × 10−3 sec⁡−1K = 2.31\, \times \,{10^{ - 3}}\,{\sec ^{ - 1}}
To find;
the time taken for concentration of A to become0.25 M0.25\,M
from the rate law expression, rate=k[A]12[B]12 = k{\left[ A \right]^{\frac{1}{2}}}{\left[ B \right]^{\frac{1}{2}}}
so, the order of the reaction is 12+12=1\dfrac{1}{2} + \dfrac{1}{2} = 1
So, reaction is of first order reaction.
Time taken when the concentration becomes 0.5M0.5M;it is the half-life period and after another half life period its concentration becomes 0.25 M0.25\,M.So overall it requires two half time.
∴Time taken=2×t12\therefore Time\,taken = 2 \times {t_{\frac{1}{2}}}
t12=0.693K ⇒t12=0.6932.31×10−3 ⇒t12=300 sec⁡ {t_{\frac{1}{2}}} = \dfrac{{0.693}}{K}\\\ \Rightarrow {t_{\frac{1}{2}}} = \dfrac{{0.693}}{{2.31 \times {{10}^{ - 3}}}}\\\ \Rightarrow {t_{\frac{1}{2}}} = 300\,\sec
Total time taken will be =2×300=600 sec⁡= 2 \times 300 = 600\,\sec.

Hence, the correct option is B.

Note: It should be noted that order of the reaction is purely an experimental quantity which is determined by various methods like use of differential rate expressions, integral rate expressions, half-life method and isolation method.