For a chemical reaction $A + 3 B \to P r o d u c t$ , it was observed that the rate of reaction increases nine times when the concentration of B increased three times by keeping the concentration of A as constant. On doubling the concentration of both rate increases eight times. Differential rate equation can be given as:

r = k [A] [B]^{3}

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r = k [A] [B]^{2}

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r = k [A]^{2} [B]

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r = k [A]^{2} [B]^{\frac{1}{3}}

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Solution

The correct option is B $r = k [A] [B]^{2}$
According to question,

$r a t e = k [A]^{m} [B]^{n}$ ------ 1

$r a t e_{1} = 9 r a t e = k [A]^{m} [3 B]^{n}$ ----------- 2

$r a t e_{2} = 8 r a t e = k [2 A]^{m} [2 B]^{n}$ --------- 3

From 1 and 2 equation we get, $n = 2.$

From 1 and 3 equation after substituting $n$ value we get, $m = 1.$

Hence $m = 1, n = 2$ .

Option B is correct.

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The correct option is B r=k[A][B]2According to question,rate=k[A]m[B]n ------ 1 rate1=9rate=k[A]m[3B]n ----------- 2rate2=8rate=k[2A]m[2B]n --------- 3From 1 and 2 equation we get, n=2.From 1 and 3 equation after substituting n value we get, m=1.Hence m=1,n=2.Option B is correct.