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Standard XII

Chemistry

Rate Constant

The rate cons...

Question

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 $(E_{a})$ of the reaction?

1842.4R

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921.2 R

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460.6 R

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230. 3 R

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Solution

The correct option is B 921.2 R
$l n \frac{k_{2}}{k_{1}} = \frac{E a}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$

$l n [\frac{10 k_{1}}{k_{1}}] = \frac{E_{a}}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$

$E a = l n \frac{10 \times 400 \times 200}{200}$

$E_{a} = 921.2 R$

Hence, option B is correct.

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Similar questions

Q. The rate constant of a reaction at temperature 200

K

is10 times less than the rate constant at 400

K

. What is theactivation energy

(E_{a})

of the reaction

(R =

gas constant)?

Q. The rate constant of a first order reaction at

25^{o} C

{0.24 s}^{- 1}

. If the energy of activation of the reaction is

{88 k J m o l}^{- 1}

, at what temperature would this reaction have rate constant of

{4 \times 10}^{- 2} s^{- 1}

(283.7 K) ?

Q. At

400 K

, the rate constant of a chemical reaction is

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

and at

300 K

, the rate constant is

6 \times 10^{- 5} s^{- 1}

. Find the value of

E_{a}

i.e. activation energy required for the reaction.
Take:

{log}_{10} (0.2) = 0.7

Q. If

k_{1} =

rate constant at temperature

T_{1}

and

k_{2} =

rate constant at temperature

T_{2}

for a first-order reaction, then which of the following relations is correct?
[

E_{a} =

activation Energy]

Q. The activation energy for a certain reaction is

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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 (Ea) of the reaction?

The correct option is B 921.2 Rlnk2k1=EaR[1T1−1T2]ln[10 k1k1]=EaR[1T1−1T2]Ea=ln10×400×200200Ea=921.2RHence, option B is correct.

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 $(E_{a})$ of the reaction?

The correct option is B 921.2 R
$l n \frac{k_{2}}{k_{1}} = \frac{E a}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$

$l n [\frac{10 k_{1}}{k_{1}}] = \frac{E_{a}}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$

$E a = l n \frac{10 \times 400 \times 200}{200}$

$E_{a} = 921.2 R$

Hence, option B is correct.