When constructing and analyzing the graphical plot of versus utilized standardly to interrogate Arrh — Chemical Kinetics Chemistry Question

💡 Solution & Explanation

The master equation is $\ln k = -\frac{E_a}{R} \frac{1}{T} + \ln A$. This matches $y = mx + c$. Therefore, slope $m = -E_a/R$ (A is correct) and intercept $c = \ln A$ (B is correct). A high $E_a$ produces a large negative value for $-E_a/R$, driving a steep drop (C is correct). If $E_a = 0$, slope $= 0$, creating a constant $y$, which plots horizontally (D is correct).