Question

2)If rate constant of reaction becomes twice when temperature increases from 300 K to 310 K, then activation energy (in Joules) of the reaction is (Given : log 2 = 0.30)

2.303 x 90 R

2.303 x 31 R

2.303 x 90 × 31

2.303 x 90 x 31 R​​

Answer 1

Answer : Given, K2/K1 = 10, T2= 310K T1= 300K, log 2= 0.30, Using arrhenius Equation, the value of activation energy can be calculated & we can solve the above problem.

Explanation : According to arrhenius Equation,

log (k2/k1 = (Ea/2.303R)×(1/T1 -1/T2)

Or, log 2 = (Ea/2.303R) × {(310-300)/(300*310)}

Or, Ea = log 2 * 2.303R * 31 *300

OOr, Ea = 0.30 * 2.303 R * 31 * 300

Or, Ea = 2.303 R × 90 × 31

Hence, the correct answer is 2.303 × 90 × 31R

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Answer 2

Answer: Activation energy is given by 2.303 × 90 × 31 R  Joules  

Explanation:

Rate constant of a reaction is given by Arrhenius equation :

$K = Ae^{-\frac{E_a}{RT} }$

K ⇒ Rate constant

A ⇒ Pre-exponential factor

Eₐ ⇒ Activation energy of reaction

T ⇒ Temperature at which reaction takes place

$K = Ae^{-\frac{E_a}{RT} }$

㏑(K) = ㏑(A) - ( Eₐ/RT )

$ln(K_1) = lnA - \frac{E_a}{RT_1}$

$ln(K_2) = lnA - \frac{E_a}{RT_2}$

$ln(K_1) - ln(K_2) = \frac{E_a}{RT_2} - \frac{E_a}{RT_1}$

Let T₁ = 300 Kelvin

Rate constant at 300 Kelvin = K₁

T₂ = 310 Kelvin

Rate constant at 310 Kelvin = K₂ = 2K₁

Therefore,

$ln(K_1) - ln(2K_1) = \frac{E_a}{310R} - \frac{E_a}{300R}$

$- ln(2) = - \frac{E_a}{R} (\frac{10}{300*310} )$

Eₐ = 30 × 310 × ㏑2 × R

Eₐ = 30 × 310 × 2.303 × ㏒2 × R

Eₐ = 30 × 310 × 2.303 × 0.30 × R

Eₐ = 90 × 31 × 2.303 × R  Joules

Eₐ =  2.303 × 90 × 31 R  Joules  

Correct option is option D.

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