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Motivation

Consider the ...

Question

Consider the following plots of rate constant versus $\frac{1}{T}$ for four different reactions. Which of the following orders is correct for the activation energies of these reactions?

E_{c} > E_{a} > E_{d} > E_{b}

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E_{b} > E_{d} > E_{c} > E_{a}

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E_{a} > E_{c} > E_{d} > E_{b}

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E_{b} > E_{a} > E_{d} > E_{c}

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Solution

The correct option is A $E_{c} > E_{a} > E_{d} > E_{b}$
Denoting activation energy by $E_{x}$
$K = A e^{\frac{- E_{x}}{R T}}$
$log K = log A - \frac{E_{x}}{2.303 R T} . . . . . (1)$
Here, the graph given in the question is of a straight line and we know that the equation of straight line is
$y = m x + c . . . . (2)$
Comparing equation 1 with 2 we get,
$Slope = \frac{- E_{x}}{2.303 R}$
So, from the graph we can conclude that the line with the most negative slope will have the maximum activation energy value.
The correct order will be
$E_{c} > E_{a} > E_{d} > E_{b}$

Suggest Corrections

Similar questions

\frac{1}{T}

l o g (k)

\frac{1}{T}

(E)

k_{1} =

T_{1}

k_{2} =

T_{2}

E_{a} =

2 A \to

c o n c e n t r a t i o n^{- 1}

t i m e^{- 1}

t_{1 / 2}

1 / T

400

10^{- 5}

s^{- 1}

500

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