Animated Solution for Chemistry - Chemical Kinetics: 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?

Activation Energy and Slopes

When dealing with chemical kinetics, the Arrhenius equation is our master key for unlocking the relationship between the rate constant ($k$), temperature ($T$), and activation energy ($E_a$). In this problem, we are presented with a graph of $\log k$ versus $\frac{1}{T}$ for four different reactions, and we need to determine the correct order of their activation energies.

Let's start by writing down the Arrhenius equation:

To make this equation match our graph, we need to take the logarithm (base 10) on both sides. This transforms our exponential relationship into a linear one:

Now, let's compare this to the standard equation of a straight line, $y=mx+c$.

Here, our y-axis variable is $\log k$, and our x-axis variable is $\frac{1}{T}$.

This means the slope of our line, $m$, is given by:

Notice the negative sign! Because the activation energy ($E_a$) and the gas constant ($R$) are always positive, the slope will always be negative. This perfectly matches our graph, where all four lines are sloping downwards.

We can rearrange our slope equation to solve for the activation energy:

This tells us something crucial: the activation energy is directly proportional to the magnitude of the slope. A steeper downward line has a more negative slope (a larger absolute value), which corresponds to a higher activation energy.

Let's look at the four lines on our graph: $a,b,c,$ and $d$. We need to rank them by their steepness.

Line $c$ is clearly the steepest, meaning it has the most negative slope. Line $a$ is the next steepest, followed by line $d$. Finally, line $b$ is the flattest, meaning it has the least negative slope.

Comparing the magnitudes of their slopes, we get:

Since a larger magnitude of slope means a higher activation energy, the order of activation energies must be:

This perfectly matches option (d). Always remember, in an Arrhenius plot of $\log k$ vs $\frac{1}{T}$, the steeper the drop, the higher the energy barrier!