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  • The rate law for a reaction between the substance A  & B is given by <img alt="\mathrm{rate = K[A]^{n}[B]^m}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%

The rate law for a reaction between the substance A  & B is given by

\mathrm{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 reaction will be as:

Option: 1

2^{n-m}


Option: 2

\frac{1}{2}\left ( m+n \right )


Option: 3

\left ( m+n \right )


Option: 4

\left ( n-m \right )


Answers (1)

best_answer
Given, expression for rate

 r=k[A]^{n}[B]^{m}

\\r'=k[2A]^{n}[\frac{B}{2}]^{m}\: \: \\\\\therefore r'=2^{\left ( n-m \right )}[A]^{n}[B]^{m}

\\\Rightarrow r'=2^{\left ( n-m \right ) } r\\\\ \Rightarrow \frac{r'}{r}=2^{\left ( n-m \right ) }

Therefore, option(1) is correct

Posted by

jitender.kumar

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