Question

Consider the following plots of $log\left(k\right)$ versus $\frac{1}{T}$ for three different reactions. Which of the following reaction has highest activation energy $\left(E\right)$?
(where, k is the rate constant)

• A. $E_a$
• B. $E_b$
• C. $E_c$
• D. All have equal activation energies.

Answer 1

The correct option is A $E_a$

For a reaction, relationship between temperature and rate constant follows the equation,
$k=A{e}^{-\frac{E_a}{RT}}...eqn\left(i\right)$
where,
$k$ : Rate constant
$E_a$ : Activation energy (in J/mol)
$R$ : Gas constant
$T$ : Temperature of reaction (in Kelvin(K))
$A$ : Arrhenius factor or frequency factor or pre-exponential factor. It is constant, specific to a particular reaction.
Taking $l{n}_e$ both sides in eqn(i)
$lnk=lnA-\frac{E_a}{RT}$
Comparing with the equation above,
$\text{Slope}\propto E_a$
All of the above graphs have a negative slope.
The reaction for which slope is the most negative is the one having highest activation energy.
Since graph a has the most negative slope , so $E_a$ is highest