The Arrhenius equation relates the rate constant (k) of a reaction to temperature (T) and activation energy (E_a). The equation is:

$k=A{e}^{-\frac{E_a}{RT}}$

Taking the natural logarithm on both sides, we get:

$\ln k=\ln A-\frac{E_a}{RT}$

This can be rewritten as:

$\ln k=-\frac{E_a}{R}\times \frac{1}{T}+\ln A$

Comparing this with the standard equation of a straight line, y = mx + c, we see that a plot of lnk vs 1/T gives a straight line with:

• Slope (m) = $-\frac{E_a}{R}$
• Intercept (c) = lnA

However, the question states the plot is lnk vs 1/(RT). Let's adjust our equation for this.

Note that 1/(RT) = (1/R) × (1/T). Substituting into the Arrhenius equation:

$\ln k=-\frac{E_a}{R}\times \frac{1}{T}+\ln A=-E_a\times \left(\frac{1}{RT}\right)+\ln A$

Now, this is in the form y = mx + c, where:

• y = lnk
• x = 1/(RT)
• Slope (m) = -E_a
• Intercept (c) = lnA

The question states that the gradient (slope) of this plot is (–y) unit. Therefore:

$m=-E_a=-y$

Solving for the activation energy E_a:

$E_a=y$

Thus, the activation energy is y unit.

Related Topics and Formulae

Arrhenius Equation: The fundamental equation $k=A{e}^{-\frac{E_a}{RT}}$ describes how the rate constant k changes with temperature T. A is the pre-exponential factor (frequency factor), E_a is the activation energy, and R is the universal gas constant.

Graphical Determination: The logarithmic form $\ln k=-\frac{E_a}{R}\times \frac{1}{T}+\ln A$ is used to determine E_a and A from experimental data. A plot of lnk vs 1/T yields a straight line with slope = -E_a/R and intercept = lnA. The question uses a modified x-axis of 1/(RT), which changes the slope to -E_a directly.