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Arrhenius Equation

Two reactions...

Question

Two reactions of the same order have equal pre-exponential factors but their activation energies differ by $41.9 J / m o l$ . Then the ratios between rate constants of these reactions at $600$ $K$ is:

0.002

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0.02

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0.2

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none of these

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Solution

The correct option is A 0.002
According to the Arrhenius equation, $K = A e^{- E a / R T}$

Where A is proportionality constant, Ea is activation energy, R is universal gas constant and T is the temperature in kelvin.

Taking log both sides,

We get, logk = logA - Ea/2.303RT

So, for 1st reaction , logk₁ = logA - Ea₁/2.303RT

For 2nd reaction, logk₂ = logA - Ea₂/2.303RT

Subtracting equation (2) from equation (1),

logk₁ - logk₂ = (Ea₁ - Ea₂)/2.303RT

Here, (Ea₁ - Ea₂) = 41.9kJ

R = 8.3 J/K/mol

And T = 600K

$l o g (\frac{k ₁}{k ₂}) = \frac{(41.9 \times 1000)}{2.303 \times 8.3 \times 600}$ = 3.66

$\frac{k ₁}{k ₂}$ = antilog(3.66)
$= 0.002$

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Two reactions of the same order have equal pre-exponential factors but their activation energies differ by 41.9 J/mol. Then the ratios between rate constants of these reactions at 600 K is:

Two reactions of the same order have equal pre-exponential factors but their activation energies differ by $41.9 J / m o l$ . Then the ratios between rate constants of these reactions at $600$ $K$ is: