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Standard XII

Chemistry

Derivation of Kp and Kc

The rate law ...

Question

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

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 C $2^{n - m}$
Earlier rate = $k [A]^{n} [B]^{m}$
New rate = $k [2 A]^{n} {(\frac{1}{2} B)}^{m} \frac{N e w R a t e}{E a r l i e r R a t e} = \frac{2^{n} A^{n} B^{m} {.2}^{- m}}{A^{n} B^{m}} = 2^{n} {.2}^{- m} = 2^{n - m}$

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Q. The rate law for a reaction between substances A and B is given by

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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 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

The correct option is C 2n−mEarlier rate = k[A]n[B]m New rate = k[2A]n(12B)mNew RateEarlier Rate=2nAnBm.2−mAnBm=2n.2−m =2n−m

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

The correct option is C $2^{n - m}$
Earlier rate = $k [A]^{n} [B]^{m}$
New rate = $k [2 A]^{n} {(\frac{1}{2} B)}^{m} \frac{N e w R a t e}{E a r l i e r R a t e} = \frac{2^{n} A^{n} B^{m} {.2}^{- m}}{A^{n} B^{m}} = 2^{n} {.2}^{- m} = 2^{n - m}$