Question

Find out the percentage of the reactant molecules crossing over the activation energy barrier at $325K$, given that $\Delta {H_{325}} = 0.12kcal,{E_{a(b)}} = + 0.02kcal$
A. $0.80\% $
B. $80.63\% $
C. $20\% $
D. None of these

Answer 1

Hint: Activation energy is the energy that must be provided to compounds to result in a chemical reaction. The activation energy (\[{E_a}\] ) of a reaction is measured in joules per mole (\[J/mol\] ), kilojoules per mole (\[kJ/mol\] ) or kilocalories per mole (\[kcal/mol\] ). Activation energy can be thought of as the magnitude of the potential barrier (sometimes called the energy barrier) separating minima of the potential energy surface pertaining to the initial and final thermodynamic state.

Complete step by step answer:
The enthalpy of formation of a reaction is the difference of activation energy of the forward and the backward reaction. Mathematically, this can be written as:
$\Delta H = {E_{a(f)}} - {E_{a(b)}}$ ….(i)
As per the question, enthalpy of formation = $\Delta {H_{325}} = 0.12kcal$
Activation energy of the backward reaction = ${E_{a(b)}} = + 0.02kcal$
Thus, applying the values in equation (i), we have the activation energy of forward reaction as:
${E_{a(f)}} = 0.12 + 0.02 = 0.14kcal/molK = 140 cal/molK$
Now, only those molecules will cross the activation energy barrier whose energy is greater than the activation energy of the forward reaction.
The formula for the fraction of molecules whose energy is greater than the activation energy is as follows:
$x = {e^{ - \left( {\dfrac{{{E_a}}}{{RT}}} \right)}}$ …. (ii)
Where, $x = $ fraction of molecules crossing the energy barrier
${E_a} = $Activation energy$ = 140cal/molK$
$R = $Universal gas constant $ = 2cal/molK$
$T = 325K$
Substituting the values in the equation (ii), we have:
$x = {e^{ - \left( {\dfrac{{140}}{{2 \times 325}}} \right)}} = {e^{ - 0.2154}}$
Taking the natural log on both the side of the equation, we have:
$ \Rightarrow \ln x = - 0.2154$
$ \Rightarrow \log x = \dfrac{{ - 0.2154}}{{2.303}} = - 0.0935$
Taking antilog on both the sides, we have:
$x = anti\log \left( { - 0.0935} \right)$
Thus, the value $x$ is found to be:
$x = 0.8063$
The percentage of molecules that cross the barrier are = $80.63\% $ .

So, the correct answer is Option B.

Note:
In some cases, rates of reaction decrease with increasing temperature. When following an approximately exponential relationship so the rate constant can still be fit to an Arrhenius expression, this results in a negative value of \[{E_a}\] . Elementary reactions exhibiting these negative activation energies are typically barrierless reactions, in which the reaction proceeding relies on the capture of the molecules in a potential well.