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Standard XII

Chemistry

Activation Energy

The rate of a...

Question

The rate of a particular reaction triples when temperature changes from $50^{o}$ C to $100^{o}$ C. What is the activation energy of the reaction (in $J . m o l^{- 1}$ )? $(l o g 3 = 0.4771, R = 8.314 K^{- 1} m o l^{- 1})$

24.012 \times 10^{- 3}

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24.012 \times 10^{3}

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22.012 \times 10^{- 3}

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22.012 \times 10^{3}

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Solution

The correct option is C $22.012 \times 10^{3}$
From Arrhenius equation we have,

$2.303 l o g \frac{K_{2}}{K_{1}} = \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$

$2.303 l o g \frac{3}{1} = \frac{E_{a}}{8.314} (\frac{1}{323} - \frac{1}{373})$

$\Rightarrow E_{a} = 22.012 \times 10^{3} J$

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Similar questions

Q. The rate of a reaction triples when temperature changes from

20^{o} C

50^{o} C

. Calculate energy of activation for the reaction

(R = 8.314 {JK}^{- 1} {mol}^{- 1}, l o g_{10} 3 = 0.477) .

(Give your answer to the nearest integer value)

Q. The rate of a reaction triples when temperature changes from

20^{o} C t o 50^{o} C .

The energy of activation for the reaction is
(

R = 8.314 J K^{- 1} m o l^{- 1}

)

(a) What is rate of reaction? Write two factors that affect the rate of reaction.

(b) The rate constant of a first order reaction increases from $4 \times 10^{- 2} t o 8 \times 10^{- 2}$ when the temperature changes from $27^{\circ} C$ to $37^{\circ} C$ . Calculate the energy of activation $(E_{a})$ . (log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021

Q. The rate of a reaction triples when temperature changes from

20^{o} C

50^{o} C

. Calculate energy of activation for the reaction.

(R = 8.314 J K^{- 1} {m o l}^{- 1})

Q. A reaction has an activation energy of

209 k J . m o l^{- 1}

. The rate increases

10

- fold when the temperature is increased from

27^{o}

C to

X^{o} C

. The temperature X is closest to.[Gas constant,

R = 8.314

J . m o l^{- 1} K^{- 1}

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The rate of a particular reaction triples when temperature changes from 50oC to 100oC. What is the activation energy of the reaction (in J.mol−1)? (log3=0.4771,R=8.314K−1mol−1)

The correct option is C 22.012×103From Arrhenius equation we have,2.303 logK2K1=EaR(1T1−1T2)2.303 log31=Ea8.314(1323−1373)⇒Ea=22.012×103 J

The rate of a particular reaction triples when temperature changes from $50^{o}$ C to $100^{o}$ C. What is the activation energy of the reaction (in $J . m o l^{- 1}$ )? $(l o g 3 = 0.4771, R = 8.314 K^{- 1} m o l^{- 1})$

The correct option is C $22.012 \times 10^{3}$
From Arrhenius equation we have,

$2.303 l o g \frac{K_{2}}{K_{1}} = \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$

$2.303 l o g \frac{3}{1} = \frac{E_{a}}{8.314} (\frac{1}{323} - \frac{1}{373})$

$\Rightarrow E_{a} = 22.012 \times 10^{3} J$