Solution:
Solution
We have Arrhenius equation for calculation of energy of activation of reaction with rate constant $$K$$ and temperature $$T$$ is
$$K=A$$ $$e^{-Ea/RT}$$ $$- (i)$$
where, $$Ea$$= Arrhenius activation energy
$$A$$= pre exponential factor
$$R$$= Ideal gas constant
According to question,
For reaction $$A$$ :-
$$T_1=200K, T_2=220K$$
$$3(r_1)_A=(r_2)_A$$
where, $$(r_1)_A$$= rate of reaction $$A$$ at $$T_1 K$$
$$(r_2)_A=$$ rate of reaction $$A$$ at $$T_2K$$
$$\because$$ Rate of reaction $$\propto$$ Rate constant
$$\Rightarrow$$ $$3(K_1)_A=(K_2)_A$$
where, $$K_1$$ and $$K_2$$ are rate constants at $$T_1$$ and $$T_2K$$
Now, $$\cfrac {(K_1)_A}{(K_2)_A}=\cfrac {1}{3}$$ $$- (ii)$$
Using $$(i)$$ in $$(ii)$$ and putting the values
$$\cfrac {(K_1)_A}{(K_2)_A}=\cfrac {1}{3}$$
$$\Rightarrow \cfrac {1}{3}= \cfrac {A\quad e^{-Ea/RT_1}}{A\quad e^{-Ea/RT_2}}$$
$$\Rightarrow \cfrac {1}{3}=$$$$ e^{-Ea/R \left(\cfrac {1}{T_2}-\cfrac {1} {T_1}\right)}$$
$$= e^{-Ea/R \left(\cfrac {1}{220}- \cfrac {1}{200}\right)}$$
$$\Rightarrow\cfrac {1}{3}= e^{+\cfrac {E_a}{R} \times \cfrac {20}{220\times 200}}$$
$$\Rightarrow \log \left( \cfrac {1}{3}\right)=\cfrac {E_a}{R}\times \cfrac {20}{220\times 200}$$
$$E_a= 2200$$ $$R$$ $$\log (1/3)$$ $$- (A)$$
For reaction $$B$$:-
$$T_1=200K$$, $$T_2=220K$$
$$\cfrac {(K_1)_B} {(K_2)_B}=\cfrac {1}{9}$$
By following the similar procedure as above,
$$\log \left(\cfrac {1}{9}\right)=\cfrac {E_b}{R}\times \cfrac {1}{2200}$$
where, $$E_b=$$ activation energy of reaction $$B$$.
$$\Rightarrow E_b=2200$$ $$R$$ $$\log\left(\cfrac {1}{9}\right)$$ $$- (B)$$
Now, from $$(A)$$ and $$(B)$$:-
$$\cfrac {E_a}{E_b}=\cfrac {log (1/3)}{log (1/9)}=\cfrac {1}{2}$$
$$\Rightarrow 2E_a=E_b$$
$$\Rightarrow 2E_A=E_B$$