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Standard XII

Chemistry

Rate of Reaction

The rate law ...

Question

The rate law for a reaction between substances A and B is given by $R a t e = k [A]^{n} [B]^{m}$ 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:

m+n

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n-m

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2^{n - m}

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\frac{1}{2^{m + n}}

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Solution

The correct option is B $2^{n - m}$
$r a t e = k {[A]}^{n} {[B]}^{m}$

On doubling the concentration of A and making the volume of B half, the new rate becomes $2^{n} \times (\frac{1}{2})^{m}$ = $2^{n - m}$ times the earlier rate.

So Ratio is $2^{n - m}$

Option C is correct.

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For the reaction : $a A + b B \to$ Products, the rate law is given by Rate = $k [A]^{m} [B]^{n}$ . On making the concentration of A two-fold and halving that of B, the ratio of the new rate to the earlier rate of the reaction would be:

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The rate law for a reaction between 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 reaction will be:

The correct option is B 2n−mrate=k[A]n[B]mOn doubling the concentration of A and making the volume of B half, the new rate becomes 2n×(12)m = 2n−m times the earlier rate.So Ratio is 2n−mOption C is correct.

The rate law for a reaction between substances A and B is given by $R a t e = k [A]^{n} [B]^{m}$ 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:

The correct option is B $2^{n - m}$
$r a t e = k {[A]}^{n} {[B]}^{m}$

On doubling the concentration of A and making the volume of B half, the new rate becomes $2^{n} \times (\frac{1}{2})^{m}$ = $2^{n - m}$ times the earlier rate.

So Ratio is $2^{n - m}$

Option C is correct.