Question

For the reaction A + B → products, it is observed that:
(1) On doubling the initial concentration of A only, the rate
of reaction is also doubled and
(2) On doubling the initial concentrations of both A and B,
there is a change by a factor of 8 in the rate of the reaction.
The rate of this reaction is given by:
(a) rate = k [A] [B]² (b) rate = k [A]² [B]²
(c) rate = k [A] [B] (d) rate = k [A]² [B]

Answer 1

Answer:

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Answer 2

Therefore the rate of the equation is $K[A].[B]^2$

Explanation:

Given reaction is

A+B→product

Let the rate of the reaction be

$K[A]^a.[B]^b$

Where K is rate constant. a and b are order of the reaction A and B respectively.

Since  on doubling the initial concentration of A only, the rate of reaction is also doubled.

Therefore the order of A is a=1

Rate= $K[A]^1.[B]^b$........(1)

Again given, on doubling the initial concentrations of both A and B, then

New rate = $K[2A]^1[2B]^b$.........(2)

There is a change by a factor of 8 in rate of the reaction.

$\frac{\textrm{new rate}}{rate} =8$

Equation (2) ÷ Equation (1)

$\frac{\textrm{new rate}}{rate} =\frac{K[2A]^1[2B]^b}{K[A]^1.[B]^b}$

$\Rightarrow 8 = 2^12^b$

$\Rightarrow 2^3=2^{1+b}$

⇒3=1+b

⇒b=2

Therefore the rate of the equation is $K[A].[B]^2$