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Arrhenius Equation

On increasing...

Question

On increasing temperature from 200K to 220K, rate constant of reaction A increases by 3 times and rate constant of reaction B increases by 9 times then correct relationship between activation energy of A and B is:

A

E_{A} = 3 E_{B}

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B

3 E_{A} = E_{B}

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C

E_{B} = 2 E_{A}

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D

E_{A} = 2 E_{B}

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Solution

The correct option is C $E_{B} = 2 E_{A}$
We have Arrhenius equation for calculation of energy of activation of reaction with rate constant $K$ and temperature $T$ is
$K = A$ $e^{- E a / R T}$ $- (i)$
where, $E a$ = Arrhenius activation energy
$A$ = pre exponential factor
$R$ = Ideal gas constant
According to question,
For reaction $A$ :-
$T_{1} = 200 K, T_{2} = 220 K$
$3 (r_{1})_{A} = (r_{2})_{A}$
where, $(r_{1})_{A}$ = rate of reaction $A$ at $T_{1} K$
$(r_{2})_{A} =$ rate of reaction $A$ at $T_{2} K$

$∵$ 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 $T_{2} K$

Now, $\frac{(K_{1})_{A}}{(K_{2})_{A}} = \frac{1}{3}$ $- (i i)$
Using $(i)$ in $(i i)$ and putting the values

$\frac{(K_{1})_{A}}{(K_{2})_{A}} = \frac{1}{3}$
$\Rightarrow \frac{1}{3} = \frac{A e^{- E a / R T_{1}}}{A e^{- E a / R T_{2}}}$
$\Rightarrow \frac{1}{3} =$ $e^{- E a / R (\frac{1}{T_{2}} - \frac{1}{T_{1}})}$
$= e^{- E a / R (\frac{1}{220} - \frac{1}{200})}$
$\Rightarrow \frac{1}{3} = e^{+ \frac{E_{a}}{R} \times \frac{20}{220 \times 200}}$
$\Rightarrow log (\frac{1}{3}) = \frac{E_{a}}{R} \times \frac{20}{220 \times 200}$
$E_{a} = 2200$ $R$ $log (1 / 3)$ $- (A)$

For reaction $B$ :-
$T_{1} = 200 K$ , $T_{2} = 220 K$
$\frac{(K_{1})_{B}}{(K_{2})_{B}} = \frac{1}{9}$

By following the similar procedure as above,
$log (\frac{1}{9}) = \frac{E_{b}}{R} \times \frac{1}{2200}$
where, $E_{b} =$ activation energy of reaction $B$ .
$\Rightarrow E_{b} = 2200$ $R$ $log (\frac{1}{9})$ $- (B)$
Now, from $(A)$ and $(B)$ :-
$\frac{E_{a}}{E_{b}} = \frac{l o g (1 / 3)}{l o g (1 / 9)} = \frac{1}{2}$
$\Rightarrow 2 E_{a} = E_{b}$
$\Rightarrow 2 E_{A} = E_{B}$

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Similar questions

Q. On increasing temperature from

200 K

220 K

rate of reaction

A

3

times and rate of reaction

B

9

times then correct relationship between activation energy of

A

B

Q. Two reactions,

A \to p r o d u c t s

B \to p r o d u c t s

, have rate constant

K_{a}

K_{b}

at temperature T and activation energies

E_{a}

E_{b}

respectively. If

K_{a} > K_{b}

E_{a} > E_{b}

and assuming that A for both the reactions is same then:

Q. Two reactions,

A \to

B \to

Products, have rate constants

k_{A}

k_{B}

T

and activation energies

E_{A}

E_{B}

respectively. If

k_{A} > k_{B}

E_{A} < E_{B}

and assuming that collision frequency for both the reactions is same, then:

Consider the Arrhenius equation given below and mark the correct option.
$k = A . e - \frac{E_{a}}{R T}$
(a) Rate constant increases exponentially with increasing activation energy and decreasing temperature
(b) Rate constant decreases exponentially with increasing activation energy and decreasing temperature
(c) Rate constant increases exponentially with decreasing activation energy and decreasing temperature
(d) Rate constant increases exponentially with decreasing activation energy and increasing temperature

Q. The rate constant of a reaction at temperature 200 K is 10 times less than the rate constant at 400K. What is the activation energy

(E_{a})

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