Chemical Kinetics and Nuclear ChemistryMedium
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
Options
A.1/2m + n
B.(m + n)
C.(n − m)
D.2(n − m)
Solution
$\frac{r_{2}}{r_{1}} = \frac{K\left( 2\lbrack A\rbrack \right)^{n}.\left( \lbrack B\rbrack/2 \right)^{m}}{K\lbrack A\rbrack^{n}.\lbrack B\rbrack^{m}} = 2^{n - m}$
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