JEE PYQ: Chemical Kinetics Question 19
Question 19 - 2020 (05 Sep Shift 2)
The rate constant ($k$) of a reaction is measured at different temperatures ($T$), and the data are plotted in the given figure. The activation energy of the reaction in kJ mol$^{-1}$ is:
[Graph shows $\ln k$ vs $\frac{10^3}{T}$ with slope $= \frac{10 - 0}{0 - 5} = -2$]
($R$ is gas constant)
(1) $2/R$
(2) $1/R$
(3) $R$
(4) $2R$
Show Answer
Answer: (4)
Solution
Arrhenius equation: $\ln k = \ln A - \left(\frac{E_a}{R \times 10^3}\right) \times \frac{10^3}{T}$. Slope of graph $= \frac{-E_a}{R \times 10^3} = \frac{-10}{5}$. $E_a = 2R \times 10^3,\text{J} = 2R,\text{kJ}$.
BETA
