JEE MAIN - Chemistry (2003 - No. 40)
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)
$${1 \over {{2^{(m + n)}}}}$$
Explanation
$$Rat{e_1} = k{\left[ A \right]^n}{\left[ B \right]^m};$$
$$Rat{e_2} = k{\left[ {2A} \right]^n}{\left[ {{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}B} \right]^m}$$
$$\therefore$$ $$\,\,\,\,{{Rat{e_2}} \over {Rat{e_1}}} = {{k{{\left[ {2A} \right]}^n}{{\left[ {{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}B} \right]}^m}} \over {k{{\left[ A \right]}^n}{{\left[ B \right]}^m}}}$$
$$ = {\left[ 2 \right]^n}{\left[ {{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}} \right]^m} = {2^n}{.2^{ - m}} = {2^{n - m}}$$
$$Rat{e_2} = k{\left[ {2A} \right]^n}{\left[ {{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}B} \right]^m}$$
$$\therefore$$ $$\,\,\,\,{{Rat{e_2}} \over {Rat{e_1}}} = {{k{{\left[ {2A} \right]}^n}{{\left[ {{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}B} \right]}^m}} \over {k{{\left[ A \right]}^n}{{\left[ B \right]}^m}}}$$
$$ = {\left[ 2 \right]^n}{\left[ {{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}} \right]^m} = {2^n}{.2^{ - m}} = {2^{n - m}}$$
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