For a reaction with an activation energy of 49-8 kJ mol -1- the ratio of the rate constant at 600 K and 300 K - K 600- K 300- is approximately- R -8-3 J mol -1 K -1correct answer -2- wrong answer -0-5 A- ln 10B- 10C- 10- eD - e 10

The correct option is D e^10Given, E_a=49.8 kJ mol^ 1 By using Arrhenius formula, K=Ae^ E_a/RT ln(K_600K_300)= E_aR(1600 1300) ln(K_600K_300)= 49.8× 10^38.3(1 ...

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Standard XII

Chemistry

Activation Energy

For a reactio...

Question

For a reaction with an activation energy of $49.8 {kJ mol}^{- 1}$ , the ratio of the rate constant at $600 K$ and $300 K$ , $\frac{K_{600}}{K_{300}}$ , is approximately:
$(R = 8.3 {J mol}^{- 1} K^{- 1})$
(correct answer +2, wrong answer -0.5)

l n (10)

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10

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10 + e

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e^{10}

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Solution

The correct option is D $e^{10}$
Given,
$E_{a} = 49.8 {kJ mol}^{- 1}$
By using Arrhenius formula,
$K = A e^{- \frac{E_{a}}{R T}}$
$l n (\frac{K_{600}}{K_{300}}) = - \frac{E_{a}}{R} (\frac{1}{600} - \frac{1}{300}) l n (\frac{K_{600}}{K_{300}}) = - \frac{49.8 \times 10^{3}}{8.3} (\frac{1}{600} - \frac{2}{600}) \Rightarrow l n (\frac{K_{600}}{K_{300}}) = \frac{49.8 \times 10^{3}}{8.3} \frac{1}{600}$
$\Rightarrow l n (\frac{K_{600}}{K_{300}}) = \frac{49.8 \times 10^{3}}{8.3 \times 600} = 10$
$\Rightarrow \frac{K_{600}}{K_{300}} = e^{10}$

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For a reaction with an activation energy of 49.8 kJ mol−1, the ratio of the rate constant at 600 K and 300 K, K600K300, is approximately: (R=8.3 J mol−1 K−1) (correct answer +2, wrong answer -0.5)

The correct option is D e10Given, Ea=49.8 kJ mol−1 By using Arrhenius formula, K=Ae−EaRT ln(K600K300)=−EaR(1600−1300)ln(K600K300)=−49.8×1038.3(1600−2600)⇒ln(K600K300)=49.8×1038.31600 ⇒ln(K600K300)=49.8×1038.3×600=10 ⇒K600K300=e10

For a reaction with an activation energy of $49.8 {kJ mol}^{- 1}$ , the ratio of the rate constant at $600 K$ and $300 K$ , $\frac{K_{600}}{K_{300}}$ , is approximately:
$(R = 8.3 {J mol}^{- 1} K^{- 1})$
(correct answer +2, wrong answer -0.5)