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The rate law of a reaction $A+B \to $ product is rate $=K [A]^n[B]^m$. On doubling the concentration of A and halving the concentration of B. Thr ratio of new rate to the earlier rate of reaction will be :

$(A)\;n-m \\ (B)\;2^{n-m} \\(C)\;\large\frac{1}{2^{m+n}} \\(D)\;m+n $

1 Answer

$r_o=K [A]^n[B]^m$
$r_1= K[2A]^n[B/2]^m$
$r_1=Ke^{n-m} [A^n][B]^m$
$r_1=r \times 2^{n-m}$
Hence b is the correct answer.
answered Apr 3, 2014 by meena.p

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