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Chemical Kinetics and Nuclear Chemistry question

2020 · 8 Jan · Shift 2 · Q4

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Chemical Kinetics and Nuclear Chemistry
2020 · 8 Jan · Shift 2 · Q4

Chemical Kinetics and Nuclear Chemistry question

2020 · 8 Jan · Shift 2 · Q4

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1

Consider the following plots of rate constant versus

{1 \over T}

for four different reactions. Which of the following orders is correct for the activation energies of these reactions? JEE Main 2020 (Online) 8th January Evening Slot Chemistry - Chemical Kinetics and Nuclear Chemistry Question 131 English

JEE Main 2020 (Online) 8th January Evening Slot Chemistry - Chemical Kinetics and Nuclear Chemistry Question 131 English

A
Ec > Ea > Ed > Eb
B
Ea > Ec > Ed > Eb
C
Eb > Ea > Ed > Ec
D
Eb > Ed > Ec > Ea

View written solutionFree

Correct answer: A

For an Arrhenius plot, the rate constant varies as

$k = A e^{-E_a/(RT)}$

Taking logarithm:

$\ln k = \ln A - \frac{E_a}{R}\left(\frac{1}{T}\right)$

So, if we plot $\ln k$ versus $1/T$ , the slope is

$\text{slope} = -\frac{E_a}{R}$

Thus, a more negative slope means a larger activation energy.

Even if the graph is shown as $k$ vs $1/T$ , the reaction whose rate constant decreases more rapidly with increase in $1/T$ corresponds to a larger activation energy.
From the given plots, curve $c$ falls most steeply with $1/T$ , then $a$ , then $d$ , and $b$ is the least steep.

Therefore,

$E_c > E_a > E_d > E_b$

Now compare with the options:

A: $E_c > E_a > E_d > E_b$ ✅
B: $E_a > E_c > E_d > E_b$ ❌
C: $E_b > E_a > E_d > E_c$ ❌
D: $E_b > E_d > E_c > E_a$ ❌

Hence, the correct option is A.

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