$$\log {{{k_2}} \over {{k_1}}} = {{{E_a}} \over {2.303R}}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)$$

Initial temperature, T₁ = 20 + 273 = 293 K

Final temperature, T₂ = 35 + 273 = 308 K

R = 8.314 JK^–1 mol^–1

As rate becomes double on raising temperature

$$ \therefore $$ r₂ = 2r₁

As rate constant, k $$\infty $$ r

k₂ = 2k₁

$$ \therefore $$ $$\log 2 = {{{E_a}} \over {2.303 \times 8.314}}\left( {{1 \over {293}} - {1 \over {308}}} \right)$$

$$ \Rightarrow $$ $$0.301 = {{{E_a}} \over {19.147}} \times {{15} \over {293 \times 308}}$$

$$ \Rightarrow $$ E_$$a$$ = 34673 J mol^–1 = 34.7 kJ mol^–1

Rate₁ = k[A]²[B]³

when concentrations of both A and B are doubled then

Rate₂ = k[2A]²[2B]³ = 32 k[A]²[B]³

$$ \therefore $$ Rate will increase by a factor of 32.

For energy 10° rise in temperature the rate of reaction doubles. So, rate = 2ⁿ
when, n = 1 rate = 2¹ = 2

when, temperature is increased from 10°C to 100°C change in temperature = 100 – 10 = 90°C

i.e., n = 9

So, rate = 2⁹ = 512 times

Rate of reaction for A + B $$ \to $$ Product

Rate = k[A]^x[B]^y …(1)

where, x and y are order w.r.t. A and B respectively. When the concentration of only B is doubled, the rate is doubled, so

R’ = k [A]^x [2B]^y = 2R …(2)

If concentration of both the reactants A and B are doubled then the rate increases by a factor of 8 so

R’’ = k[2A]^x[2B]^y = 8R ...(3)

= k2^x 2^y [A]^x [B]^y = 8R …(4)

From equation (1) and (2), we get

$${{2R} \over R} = {{{{\left[ A \right]}^x}{{\left[ {2B} \right]}^y}} \over {{{\left[ A \right]}^x}{{\left[ B \right]}^y}}}$$

$$ \Rightarrow $$ 2 = 2^y

$$ \Rightarrow $$ y = 1

From equation (1) and (4), we get

$${{8R} \over R} = {{{2^x}{2^y}{{\left[ A \right]}^x}{{\left[ B \right]}^y}} \over {{{\left[ A \right]}^x}{{\left[ B \right]}^y}}}$$

$$ \Rightarrow $$ 8 = $${{2^x}{2^y}}$$

Substituting the value of y gives

$${{2^x}{2^1}}$$ = 8

$$ \Rightarrow $$ $${{2^x}}$$ = 4

$$ \Rightarrow $$ x = 2

By replacing the values of x and y in

rate law; rate = k[A]²[B]