Chemical Thermodynamics — Part 2: Enthalpy of reaction, Hess’s law, entropy and Gibbs energy: visual lesson, CBSE Class 11 Chemistry

Lesson

Enthalpy, Entropy and Spontaneity

Part 2 — measuring heats of reaction, combining them by Hess’s law, and predicting which changes happen on their own.

In this lesson you will

Measure ΔU and ΔH by bomb and constant-pressure calorimetry
Define the standard enthalpies of formation, combustion, atomisation, sublimation, phase transition, ionisation, solution and dilution
Apply Hess’s law and bond enthalpies to find ΔH of reactions
Use ΔS = q_rev/T, ΔG = ΔH − TΔS and ΔG° = −2.303RT log K to judge spontaneity and equilibrium

Table of four questions of Part 2 and the tool used to answer each
Question	Tool
How much heat does a reaction give out?	Calorimeter
Can I get an unmeasurable ΔH from known ones?	Hess’s law
Does a reaction go by itself?	ΔG = ΔH − TΔS
How far does it go?	ΔG° and K

Concept

Measuring ΔU and ΔH: calorimetry

A bomb calorimeter is a sealed steel vessel at constant volume: it measures $\Delta U$ . A coffee-cup (open) calorimeter is at constant pressure: it measures $\Delta H$ . Heat absorbed by the calorimeter $q = C\Delta T$ or $mc\Delta T$ .

q_V = -C_\text{cal}\,\Delta T \quad\Rightarrow\quad \Delta U = \dfrac{q_V}{n}

Table comparing bomb and constant-pressure calorimeters
Feature	Bomb calorimeter	Constant-pressure calorimeter
Condition	Constant volume, strong steel vessel	Constant pressure, open to the air
Measures	$\Delta U$	$\Delta H$
Used for	Combustion of fuels and foods	Neutralisation, solution, reactions in water
Conversion	$\Delta H = \Delta U + \Delta n_g RT$	None needed

Worked example

Worked example: constant-pressure calorimetry

Problem. 50 mL of 1.0 M HCl is mixed with 50 mL of 1.0 M NaOH in a coffee-cup calorimeter; the temperature rises by 6.8 K (given). Take the 100 g of solution to have $c = 4.18\ \text{J g}^{-1}\text{K}^{-1}$ . Find $\Delta H$ per mole of water formed.

Step 1. Heat gained by solution $= mc\Delta T = 100 \times 4.18 \times 6.8 = 2842\ \text{J}$ .

Step 2. Moles of water formed $= 0.050\ \text{mol}$ .

Step 3. $\Delta H = -\dfrac{2842}{0.050} = -56850\ \text{J mol}^{-1} = -56.8\ \text{kJ mol}^{-1}$ (textbook value about −57).

Table of heat gained, moles of water and the enthalpy of neutralisation calculated from the calorimeter data
Quantity	Value
Heat gained, J	2842
Moles of water	0.050
ΔH / kJ mol⁻¹	-56.8

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