Chapter 10, Problem 56IP

Interpretation Introduction

Interpretation: Among the given reactions, those that are expected to be thermodynamically favourable have to be determined.

Concept Introduction:

Halogenation of alkanes:

The halogen atom replaces a hydrogen atom in alkane and thereby gets inserted to the alkane.

Example:

Thermodynamic consideration of halogenation of alkanes:

For a reaction to be thermodynamically favourable, the sign of $\text{ΔG}$ has to be analysed.

The expression for $\text{ΔG}$ is:

$\text{ΔG}=\text{ΔH}\text{-}\text{T}\text{ΔS}$

$\begin{array}{l}\text{Where,}\\ \text{ΔG}=\text{Change}\text{in}\text{the}\text{Gibb's}\text{free}\text{energy}.\\ \text{ΔH}=\text{Change}\text{in}\text{the}\text{enthalpy}\text{. }\\ \text{ΔS}=\text{Change}\text{in}\text{the}\text{entropy}\text{.}\end{array}$

In the halogenation reaction shown above, it can be observed as the two reactants reacted to form two products. So the number of reactants and products are being the same without any change. Hence, $\text{ΔS}$ is not needed to consider for the halogenation reaction. So, the whole term $\text{T}\text{ΔS}$ can be neglected.

Thus the expression of $\text{ΔG}$ becomes as:

$\text{ΔG}\sim \text{ΔH}$

This implies that the sign of $\text{ΔG}$ will analysed by accessing only the term $\text{ΔH}$.

• If $\text{ΔH}$ is negative, then $\text{ΔG}$ will also be negative. So the reaction will be spontaneous, exothermic and thermodynamically favourable reaction.
• If $\text{ΔH}$ is positive, then $\text{ΔG}$ will also be positive. So the reaction will be non-spontaneous, endothermic and thermodynamically unfavourable reaction.

$\Delta H$can be calculated as follows:

$\text{Δ}{\text{H}}_{\text{rxn}}\text{=}{\displaystyle \sum \text{Δ}\text{H}\left(\text{bonds}\text{broken}\right)}\text{-}{\displaystyle \sum \text{Δ}\text{H}\left(\text{bonds}\text{formed}\right)}$