Question

Consider the following graph of log$_{10}$K vs T where K is rate constant and T is temperature. The straight line BC has slope, tan$\theta$ = -$\frac{40}{2.303}$ and an intercept of 8 on Y-axis Thus E$_{a}$, the energy of activation is (in calorie)

Answer 1

80

Answer 2

The Arrhenius equation is given by $k = A e^{-E_a/RT}$. Taking the base-10 logarithm, we get: $\log_{10} K = \log_{10} A - \frac{E_a}{2.303RT}$ This equation can be rearranged to represent a straight line when $\log_{10} K$ is plotted against $1/T$: $\log_{10} K = \left(-\frac{E_a}{2.303R}\right) \left(\frac{1}{T}\right) + \log_{10} A$ The slope of this line is $m = -\frac{E_a}{2.303R}$. We are given that the slope $\tan\theta = -\frac{40}{2.303}$. Equating the slope: $-\frac{E_a}{2.303R} = -\frac{40}{2.303}$ This simplifies to: $\frac{E_a}{R} = 40$ $E_a = 40R$ To find $E_a$ in calories, we use the value of the gas constant $R \approx 2 \, cal/(mol \cdot K)$. $E_a = 40 \times 2 \, cal/mol$ $E_a = 80 \, cal/mol$ The plot of $\log_{10} K$ vs $1/T$ is linear according to the Arrhenius equation $\log_{10} K = \log_{10} A - \frac{E_a}{2.303RT}$. The slope of this line is $m = -\frac{E_a}{2.303R}$. Given $m = -\frac{40}{2.303}$, we have $\frac{E_a}{R} = 40$. Using $R \approx 2 \, cal/(mol \cdot K)$, $E_a = 40 \times 2 = 80 \, cal/mol$.