The rate law for a chemical reaction is given as , rate = K [A]²[B]. When the concentration of ‘A’ is increased by 1.414 times and concentration of B is halved then

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rate becomes half of initial rate

rate remains constant

rate becomes four times the initial rate

rate becomes two times the initial rate

answer is C.

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Detailed Solution

$Let {[A]}_{initial} = {[B]}_{initial} = 1 M; Initial rate, r_{1} = K {[A]}^{2} [B]; r_{1} = K {[1]}^{2} {[1]}^{1}; r_{1} = K; Final rate, r_{2} = K {[\sqrt{2}]}^{2} {[\frac{1}{2}]}^{1}; r_{2} = K$

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