Question

The rate law for a reaction between the substances A and B is given by
rate = $k\left[A{]}^n\right[B{]}^m$
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

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

Answer 1

Answer: (A)

$2^{(n-m)}$

Rate =

$Latusconsidertheratebecomes{x}^{y}timesidconcentrationischangedtoxtimesofthegivenreactanthavingythorderofreaction$

$Rateofreaction=k{\left[A\right]}^n{\left[B\right]}^m$

$SincetheconcentrationofAisdoubledhencex=2andy=n,thustheratebecomes$

$RateofA=2^{n}times$

$SincetheconcentrationofBisreducedtohalf,hencex=\frac{1}{2}andy=m,thustherateis$

$RateofB={\left(\frac{1}{2}\right)}^{m}times$

$Nowthevalueofnetrateofreaction=2^n\times {\left(\frac{1}{2}\right)}^{m}times={\left(2\right)}^{n-m}times$