organic chemistry - Why substitution and elimination reactions are favored respectively at lower and higher temperature?

I read about the competition between nucleophilic substitution and nucleophilic elimination depending on temperature here. Though the webpage clearly says higher temperature favors elimination while lower temperature favors substitution, the explanation on the webpage is unclear.

Need a clear explanation regarding the phenomenon.

Answer

For the nucleophilic substitutions and eliminations, we can draw up these four generalised schemes:

$$\begin{align} &\mathrm{S_N1}{:} & \ce{R-X + Nu- &-> R+ + X- + Nu- -> R-Nu + X-}\\ &\mathrm{S_N2}{:} & \ce{R-X + Nu- &-> R-Nu + X-}\\ &\mathrm{E1}{:} & \ce{RCH2-CR2-X + B- &-> RCH2-CR2+ + X- + B- -> RCH=CR2 + X- + HB}\\ &\mathrm{E2}{:} & \ce{RCH2-CH2-X + B- &-> RCH=CH2 + X- + HB}\end{align}$$

Clearly, each substitution keeps the number of freely diffusing particles the same while each elimination increases it by one. Therefore, the entropic term of the eliminations is likely to be higher than that of the substitutions. The rest is given by the Gibbs free energy equation: $$\Delta G = \Delta H - T\Delta S$$

The entropic factor scales with temperature as Bryce already mentioned so this outweighs any other effects at high temperatures.