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Reaction rate between two substance $A$ and $B$ is expressed as following: rate $=k[A]^n[B]^m$
If the concentration of $\mathrm{A}$ is doubled and concentration of $\mathrm{B}$ is made half of initial concentration, the ratio of the new rate to the earlier rate will be:
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If the concentration of $\mathrm{A}$ is doubled and concentration of $\mathrm{B}$ is made half of initial concentration, the ratio of the new rate to the earlier rate will be:
Solution:
1606 Upvotes
Verified Answer
The correct answer is:
$\left.2^{(n-m}\right)$
$\left.2^{(n-m}\right)$
$$
\text { } \begin{aligned}
\text { Rate }_1 & =k[A]^n[B]^m \\
\text { Rate }_2 & =k[2 A]^n\left[\frac{1}{2} B\right]^m \\
\therefore \frac{\text { Rate }_2}{\text { Rate }_1} & =\frac{k[2 \mathrm{~A}]^n\left[\frac{1}{2} \mathrm{~B}\right]^m}{k[\mathrm{~A}]^n[\mathrm{~B}]^m}=(2)^n\left(\frac{1}{2}\right)^m \\
& =2^n \cdot(2)^{-m}=2^{n-m}
\end{aligned}
$$
\text { } \begin{aligned}
\text { Rate }_1 & =k[A]^n[B]^m \\
\text { Rate }_2 & =k[2 A]^n\left[\frac{1}{2} B\right]^m \\
\therefore \frac{\text { Rate }_2}{\text { Rate }_1} & =\frac{k[2 \mathrm{~A}]^n\left[\frac{1}{2} \mathrm{~B}\right]^m}{k[\mathrm{~A}]^n[\mathrm{~B}]^m}=(2)^n\left(\frac{1}{2}\right)^m \\
& =2^n \cdot(2)^{-m}=2^{n-m}
\end{aligned}
$$
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