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.
When the question is asking about favorability, it is really asking about the magnitude of the Gibbs free energy. The more negative this value, the more favorable a reaction is. Gibbs depends on enthalpy and a temperature scaled entropy: $$\Delta G=\Delta H-T\Delta S$$
When we think of nucleophilic elimination, we know that a hydrogen atom is removed by a nucleophilic group resulting in our final product. Since a hydrogen is removed in elimination, while left intact during substitution, there are more final species formed during an elimination reaction. According to general chemistry principles, this means that there was a larger increase in entropy compared to the substitution. As temperature increases this larger entropy becomes more noticeable as it is now scaled by a larger $T$.
Therefore at low temperature, the enthalpy term dominates Gibbs and the substitution reaction prevails. However as $T$ grows, so does the entropy term and at high $T$ this overrides enthalpy, making the elimination reaction more favorable.
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.
Breaking bonds requires an input of energy. Elimination reactions require the breakage of more bonds than substitution reactions, and therefore require a higher input of energy. This indicates that the activation energy of elimination reactions is higher than that of substitution reactions. Increasing temperature will allow more molecules to reach the required activation energy threshold. Therefore, as temperatures is increased, elimination is favored.
