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Standard XII

Chemistry

Activation Energy

The activatio...

Question

The activation energy for a reaction at temperature T K was found to be 2.303 RT J $m o l^{- 1}$ . The ratio of the rate constant to Arrhenius factor is :

$2 \times 10^{- 2}$

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$2 \times 10^{- 3}$

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$10^{- 2}$

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$10^{- 1}$

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Solution

The correct option is D
$10^{- 1}$

$2.303 l o g K = 2.303 l o g A - \frac{E_{a}}{R T}$

Given $E_{a} = 2.303 R T$

$\Rightarrow 2.303 l o g K = 2.303 l o g A - \frac{2.303 R T}{R T}$

(or) $l o g K = l o g A - 1$

$l o g \frac{K}{A} = - 1$ (or) $\frac{K}{A} = 10^{- 1}$

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Similar questions

The activation energy for a reaction at the temperature TK was found to be 2.303 $R T J m o l^{- 1}$ . The ratio of the rate constant to Arrhenius factor is ___

Q. The activation energy for a reaction at the temperature

T

K was found to be 2.303 RT

J / m o l^{- 1}

. The ratio of the rate constant to the Arrhenius factor is:

Q. In Arrhenius equation for a certain reaction, the value of A (Arrhenius factor) and

E_{a}

(activation energy) are

10^{13} s^{- 1} and 12.471 k J m o l^{- 1}

respectively. At what temperature, the reaction will have specific rate constant of

10^{- 3} s^{- 1}

The rates of chemical reactions are strongly affected by temperature. Arrhenius (1889) gave following relation between rate constant and temperature:

k = A e^{- E_{a} / R T}

This equation is called Arrhenius equation. The constant '

A

' is called Arrhenius constant or pre-exponential factor.

The logarithm of Arrhenius equation is:

{log}_{10} k = {log}_{10} A - \frac{E_{a}}{2.303 R T}

This equation for a reaction is:

{log}_{10} k = 14 - \frac{1.25 \times 10^{4} K}{T}

Energy of activation in

k

c a l

{m o l}^{- 1}

is:

Q. The rate constant, the activation energy and the arrhenius parameter of a chemical reaction at

25^{0} C

are

3 \times 10^{- 4} s^{- 1}

, 104.4KJ / mol and

6 \times 10^{- 4} s^{- 1}

respectively. The value of the rate constant at

T \to \infty

is:

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The activation energy for a reaction at temperature T K was found to be 2.303 RT J mol−1. The ratio of the rate constant to Arrhenius factor is :

The correct option is D 10−1 2.303 log K=2.303 log A−EaRT Given Ea=2.303 RT ⇒ 2.303 log K=2.303 log A−2.303 RTRT (or) log K=log A−1 log KA=−1 (or) KA=10−1

The activation energy for a reaction at temperature T K was found to be 2.303 RT J $m o l^{- 1}$ . The ratio of the rate constant to Arrhenius factor is :

The correct option is D
$10^{- 1}$

$2.303 l o g K = 2.303 l o g A - \frac{E_{a}}{R T}$

Given $E_{a} = 2.303 R T$

$\Rightarrow 2.303 l o g K = 2.303 l o g A - \frac{2.303 R T}{R T}$

(or) $l o g K = l o g A - 1$

$l o g \frac{K}{A} = - 1$ (or) $\frac{K}{A} = 10^{- 1}$