The rate law for a reaction between the substances A and B is given by Rate = k[A]n [B]m On doubling the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be as

Question

The rate law for a reaction between the substances A and B is given by Rate = k[A]ⁿ [B]^m On doubling the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be as

(m + n)

(n - m)

2^{( n - m)}

${1 \over {{2^{(m + n)}}}}$

Solution

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