$\mathrm{A}:\mathrm{k}={\mathrm{Ae}}^{-\frac{\mathrm{Ea}}{\mathrm{RT}}}$(Arrhenius rate equation)

As $\mathrm{Ea}$(Activation energy) increases $\mathrm{k}$(rate constant) decreases

B :Higher activation energy, greater is the effect of given temperature rise on rate constant 'k'. Activation energy is the minimum energy needed for the reaction to occur. When temperature increases, Kinetic Energy also increases; as temperature increases, more molecules have higher Kinetic Energy, and thus the fraction of molecules that have high enough Kinetic Energy to overcome the energy barrier also increases.

Temperature coefficient $=\frac{{\mathrm{k}}_{\mathrm{T}+10}}{{\mathrm{k}}_{\mathrm{T}}}$

Option (C) is wrong. $\mathrm{\Delta k}$ may be greater or lesser depending on temperature.

$\mathrm{D}: \mathrm{lnk}=\mathrm{lnA}-\frac{\mathrm{Ea}}{\mathrm{RT}}$ in this equation the slope is $-\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}$.