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Question

In a reaction, A+B Product, rate is doubled when the concentration of B is doubled and rate increases by a factor of 8 when the concentrations of both the reactants (A and B) are doubled. Rate law for the reaction can be written as :


A

Rate =k[A][B]2

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B

Rate =k[A]2[B]2

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C

Rate =k[A][B]

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D

Rate =k[A]2[B]

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Solution

The correct option is D

Rate =k[A]2[B]


Let the order of reaction with respect to A and B is x and y respectively. So, the rate law can be given as
R=k[A]x[B]y …(i)
When the concentration of only B is doubled, the rate is doubled, so
R=k[A]x[2B]y=2R ….(ii)
If concentrations of both the reactants A and B are doubled, the rate increases by a factor of 8, so R′′=k[2A]x[2B]y=8R ….(iii)
k2x2y[A]x[B]y=8R ....(iv)
Form Eqs. (i) and (ii), we get
2RR=([A]x[2B]y)([A]x[B]y)2=2yy=1
From Eqs.(i) and (iv), we get
8RR=(2x2y[A]x[B]y)([A]x[B]y)or8=2x2y
Substitution of the value of y gives,
8=2x214=2x(2)2=(2)xX=2
Substitution of the value of x and y in Eq. (i) gives,
R=k[A]2[B]


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