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Half life

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

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

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Solution

The correct option is B $E_{c} > E_{a} > E_{d} > E_{b}$
To avoid confusion, in this question we’ll be 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 = mx + 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.
$E_{c} > E_{a} > E_{d} > E_{b}$

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