Consider the following plots of log of rate constant (log ) vs 1/T for three different reactions. The correct order of activation energies of these reactions is [2025]

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Q.
Consider the following plots of log of rate constant $k$ (log $k$ ) vs 1/T for three different reactions. The correct order of activation energies of these reactions is [2025]

1 $E a_{2} > E a_{1} > E a_{3}$

2 $E a_{1} > E a_{3} > E a_{2}$

3 $E a_{1} > E a_{2} > E a_{3}$

4 $E a_{3} > E a_{2} > E a_{1}$

Ans.
(1)

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

$\ln k = \ln A + \ln e^{- E_{a} / R T}$

$\ln k = \ln A - \frac{E_{a}}{R T}$

$\log k = \log A - \frac{E_{a}}{2.303 R} (\frac{1}{T})$

Slope of log $k$ vs $\frac{1}{T}$ graph is $- \frac{E_{a}}{2.303 R}$

Thus more negative slope means more activation energy.

Solution

Q.
Consider the following plots of log of rate constant $k$ (log $k$ ) vs 1/T for three different reactions. The correct order of activation energies of these reactions is [2025]

1 $E a_{2} > E a_{1} > E a_{3}$

2 $E a_{1} > E a_{3} > E a_{2}$

3 $E a_{1} > E a_{2} > E a_{3}$

4 $E a_{3} > E a_{2} > E a_{1}$

Ans.
(1)

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

$\ln k = \ln A + \ln e^{- E_{a} / R T}$

$\ln k = \ln A - \frac{E_{a}}{R T}$

$\log k = \log A - \frac{E_{a}}{2.303 R} (\frac{1}{T})$

Slope of log $k$ vs $\frac{1}{T}$ graph is $- \frac{E_{a}}{2.303 R}$

Thus more negative slope means more activation energy.

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Solution

Q. Consider the following plots of log of rate constant k (log k) vs 1/T for three different reactions. The correct order of activation energies of these reactions is [2025]

1 Ea2>Ea1>Ea3

2 Ea1>Ea3>Ea2

3 Ea1>Ea2>Ea3

4 Ea3>Ea2>Ea1

Ans. (1) k=Ae-Ea/RT lnk=lnA+lne-Ea/RT lnk=lnA-EaRT logk=logA-Ea2.303R(1T) Slope of log k vs 1T graph is -Ea2.303R Thus more negative slope means more activation energy.

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Q.
Consider the following plots of log of rate constant $k$ (log $k$ ) vs 1/T for three different reactions. The correct order of activation energies of these reactions is [2025]

1 $E a_{2} > E a_{1} > E a_{3}$

2 $E a_{1} > E a_{3} > E a_{2}$

3 $E a_{1} > E a_{2} > E a_{3}$

4 $E a_{3} > E a_{2} > E a_{1}$

Ans.
(1)

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

$\ln k = \ln A + \ln e^{- E_{a} / R T}$

$\ln k = \ln A - \frac{E_{a}}{R T}$

$\log k = \log A - \frac{E_{a}}{2.303 R} (\frac{1}{T})$

Slope of log $k$ vs $\frac{1}{T}$ graph is $- \frac{E_{a}}{2.303 R}$

Thus more negative slope means more activation energy.