Sol. Reaction (i) A → products

(ii) B → products

Both are first order reactions. The rate of reaction is doubled when temperature is raised from 300 K to 310 K.

So, = 2 .....(i)

(on the basis of rate of reaction)

At 310 K half-life period of reaction (i) is

30 minutes.

So (kA)310 = = per min

= 0.0231 ......(ii)

At 310 K, B decomposes twice as fast as ‘A’

Thus, = 2 ......(iii)

If (Ea)ii is half than (Ea)i

∴ = .....(iv)

Now using Arrhenius equation

k = Ae –Ea/RT

We get, Ea = log 10 .....(v)

From equation number (i) and (ii)

=2 or (kA)300= =0.01155 min–1

From equation no. (ii) and (iii)

= 2

or (kB)310 = = 0.0462 min–1

From equation (v) the value of (Ea)1 is calculated for reaction (i)

Hence (Ea)i = log 10 = ....(vi)

= log 10 2

From equation number (v) for reaction (ii)

(Ea)(ii) = log 10

= log 10 .....(vii)

From equation (vii) and (vi)

=

or =

log10 2 = log 10

0.5 × 0.3010 = log 10

0.15050 = log 10

or 1.415 =

(kB)300 = = 0.03265 minute–1

= 32.65 × 10–3 minute–1