Chapter 8.1, Problem 1RQ

Interpretation Introduction

Interpretation :

The driving forces that indicate that a chemical reaction is likely to occur must be explained.

Concept Introduction :

Some parameters are there which help to make a reaction possible. These are driving forces.

Answer to Problem 1RQ

Some of the driving forces for a chemical reaction are given below. These tend to make a reaction to occur.

• Formation of precipitate from two clear solutions
• Formation of water which can be removed to shift the equilibrium towards product side
• Transfer of electrons to gain more stability
• Formation of gas which can escape from the system and drives the reaction more towards product side.
• Increase of entropy, decrease in enthalpy and decrease in Gibb’s free energy.

Explanation of Solution

In the precipitation reaction concentration of the ions producing the precipitate decreases which drags the reaction more towards the product side to maintain the same equilibrium constant value.

  ${\text{NaCl(aq)+AgNO}}_{\text{3}}\text{(aq)}\to \text{AgCl(s)}\downarrow {\text{+NaNO}}_{\text{3}}\text{(aq)}\text{.}$

Here formation of AgCl(s) drives the reaction to product side.

In esterification reaction water is removed by acid catalyst to drive the reaction towards product side.

  ${\text{CH}}_{\text{3}}{\text{COOH(l)+CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH(l)}\to {\text{CH}}_{\text{3}}{\text{COOCH}}_{\text{2}}{\text{CH}}_{\text{3}}{\text{(l)+H}}_{\text{2}}\text{O(l)}\text{.}$

Few drops of concentrated sulfuric acid is used to take care of the water formed in this reaction.

  ${\text{Zn(s)+Cu}}^{\text{2+}}\text{(aq)}\to {\text{Cu(s)+Zn}}^{\text{2+}}\text{(aq)}\text{.}$

Zn being more electropositive it has a tendency to transfer electrons to Cu²⁺.

  ${\text{C(s)+O}}_{\text{2}}\text{(g)}\to {\text{CO}}_{\text{2}}\text{(g)}\text{.}$

In this reaction C is converted to CO₂ which escapes from the system and drags the reaction towards product side.

For spontaneous reaction Gibb’s free energy change (ΔG) must be negative as per thermodynamics.

  $\text{ΔG=ΔH-TΔS}\text{.}$

From this relation it is clear that when enthalpy change (ΔH) is highly negative and entropy change (ΔS) is highly positive then Gibb’s free energy change (ΔG) is highly negative which can drive a reaction to occur.