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:

$n - 1$

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$m + n$

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$2^{- (m + n)}$

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$2^{(m - n)}$

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Solution

The correct option is D
$2^{(m - n)}$

$(R a t e)_{1} = k [A]^{w} [B]^{n}; (R a t e)_{2} = k (2 [A])^{w} (\frac{[B]}{2})^{n}$

$H e n c e, \frac{(r a t e)_{2}}{(r a t e)_{1}} = 2^{m} \times (\frac{1}{2})^{n} = 2^{m - n}$

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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:

Q. 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

Q. The rate law for 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 reaction will be

Q. 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:

Q. The rate law for a reaction between the 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 the reaction will be as:

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For the reaction : aA+bB→ 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:

The correct option is D 2(m−n) (Rate)1=k[A]w[B]n;(Rate)2=k(2[A])w([B]2)n Hence,(rate)2(rate)1=2m×(12)n=2m−n

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:

The correct option is D
$2^{(m - n)}$

$(R a t e)_{1} = k [A]^{w} [B]^{n}; (R a t e)_{2} = k (2 [A])^{w} (\frac{[B]}{2})^{n}$

$H e n c e, \frac{(r a t e)_{2}}{(r a t e)_{1}} = 2^{m} \times (\frac{1}{2})^{n} = 2^{m - n}$