CIE AS/A Level Chemistry 5.2 Hess’s Law Study Notes- 2025-2027 Syllabus
CIE AS/A Level Chemistry 5.2 Hess’s Law Study Notes – New Syllabus
CIE AS/A Level Chemistry 5.2 Hess’s Law Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on AS/A Level Chemistry latest syllabus with Candidates should be able to:
Hess’s Law and Simple Energy Cycles
Hess’s Law allows enthalpy changes to be calculated indirectly. It is particularly useful when the enthalpy change of a reaction cannot be measured directly.
Hess’s Law
Hess’s Law states:
The enthalpy change of a reaction is the same regardless of the route taken, provided the initial and final states are the same.
This works because enthalpy is a state function — it depends only on the initial and final states, not the path.
Why Use Hess’s Law?
- Some reactions cannot be measured directly (e.g., carbon to carbon monoxide).
- Energy cycles allow indirect calculation using known enthalpy changes:
- \( \Delta H_\mathrm{f}^\circ \), \( \Delta H_\mathrm{c}^\circ \), or solution/neutralisation enthalpies.
Constructing Simple Energy Cycles
Two common Hess cycles:
- Formation cycle (using \( \Delta H_\mathrm{f}^\circ \))
- Combustion cycle (using \( \Delta H_\mathrm{c}^\circ \))
Energy changes are added or subtracted depending on the direction of arrows.
1. Formation Cycle
To find the enthalpy change for:
Reactants → Products
Use elements in their standard states as the reference level.
\( \Delta H_\mathrm{r}^\circ = \Sigma \Delta H_\mathrm{f}^\circ(\text{products}) – \Sigma \Delta H_\mathrm{f}^\circ(\text{reactants}) \)
2. Combustion Cycle
Products and reactants are both combusted to the same final products (usually \( \mathrm{CO_2} \) and \( \mathrm{H_2O} \)).
\( \Delta H_\mathrm{r}^\circ = \Sigma \Delta H_\mathrm{c}^\circ(\text{reactants}) – \Sigma \Delta H_\mathrm{c}^\circ(\text{products}) \)
Arrows point downwards because combustion is always exothermic.
Example
State Hess’s Law in simple terms.
▶️ Answer / Explanation
The enthalpy change of a reaction is independent of the route taken as long as the starting and ending conditions are the same.
Example
Given the following:
\( \Delta H_\mathrm{f}^\circ (\mathrm{CO_2}) = -394\ \mathrm{kJ\ mol^{-1}} \)
\( \Delta H_\mathrm{f}^\circ (\mathrm{H_2O(l)}) = -286\ \mathrm{kJ\ mol^{-1}} \)
Calculate \( \Delta H_\mathrm{r}^\circ \) for:
\( \mathrm{CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O} \)
▶️ Answer / Explanation
Products:
\( 1\times -394 + 2\times -286 = -966\ \mathrm{kJ} \)
Reactants:
\( \mathrm{CH_4} = -75\ \mathrm{kJ\ mol^{-1}} \) (given)
\( \mathrm{O_2} = 0 \) (element in standard state)
Overall:
\( \Delta H_\mathrm{r}^\circ = -966 – (-75) = -891\ \mathrm{kJ\ mol^{-1}} \)
Example
Using the combustion cycle and the following data:
\( \Delta H_\mathrm{c}^\circ(\mathrm{C_2H_6}) = -1560\ \mathrm{kJ\ mol^{-1}} \)
\( \Delta H_\mathrm{c}^\circ(\mathrm{C_2H_4}) = -1411\ \mathrm{kJ\ mol^{-1}} \)
\( \Delta H_\mathrm{c}^\circ(\mathrm{H_2}) = -286\ \mathrm{kJ\ mol^{-1}} \)
Calculate \( \Delta H_\mathrm{r}^\circ \) for: \( \mathrm{C_2H_6 \rightarrow C_2H_4 + H_2} \)
▶️ Answer / Explanation
Using Hess’s Law:
\( \Delta H_\mathrm{r}^\circ = \Sigma \Delta H_\mathrm{c}^\circ(\text{reactants}) – \Sigma \Delta H_\mathrm{c}^\circ(\text{products}) \)
Reactants: \( -1560 \)
Products: \( -1411 + -286 = -1697 \)
\( \Delta H_\mathrm{r}^\circ = -1560 – (-1697) = +137\ \mathrm{kJ\ mol^{-1}} \)
Endothermic.
Calculations Using Hess Cycles
Hess cycles allow enthalpy changes to be calculated indirectly. These calculations often use known enthalpy values such as formation, combustion, solution enthalpies, or bond energies.
(a) Determining Enthalpy Changes that Cannot Be Found Directly
Some reactions cannot be performed experimentally (e.g., incomplete combustion, unstable intermediates). Hess’s Law allows their enthalpy changes to be determined indirectly using an energy cycle.
Key formula (formation cycle):
\( \Delta H_\mathrm{r}^\circ = \Sigma \Delta H_\mathrm{f}^\circ(\text{products}) – \Sigma \Delta H_\mathrm{f}^\circ(\text{reactants}) \)
Key formula (combustion cycle):
\( \Delta H_\mathrm{r}^\circ = \Sigma \Delta H_\mathrm{c}^\circ(\text{reactants}) – \Sigma \Delta H_\mathrm{c}^\circ(\text{products}) \)
(b) Calculations Using Bond Energy Data
Bond energies can be used to estimate reaction enthalpies based on bonds broken (reactants) and bonds formed (products).
\( \Delta H_\mathrm{r} = \Sigma E_\text{broken} – \Sigma E_\text{formed} \)
- Breaking bonds is endothermic → positive values.
- Forming bonds is exothermic → negative contribution.
- Bond energy values are usually averages.
Example
Use Hess’s Law to determine the enthalpy change for a reaction that cannot be measured directly.
Given:
\( \Delta H_\mathrm{f}^\circ(\mathrm{NO_2}) = +34\ \mathrm{kJ\ mol^{-1}} \)
\( \Delta H_\mathrm{f}^\circ(\mathrm{N_2O_4}) = +9\ \mathrm{kJ\ mol^{-1}} \)
Find \( \Delta H_\mathrm{r}^\circ \) for:
\( \mathrm{N_2O_4 \rightarrow 2NO_2} \)
▶️ Answer / Explanation
Products: \( 2 \times 34 = 68\ \mathrm{kJ\ mol^{-1}} \)
Reactants: \( 9\ \mathrm{kJ\ mol^{-1}} \)
Overall:
\( \Delta H_\mathrm{r}^\circ = 68 – 9 = +59\ \mathrm{kJ\ mol^{-1}} \)
Endothermic reaction.
Example
Use a combustion cycle to determine \( \Delta H_\mathrm{r}^\circ \) for:
\( \mathrm{CO + \tfrac{1}{2}O_2 \rightarrow CO_2} \)
Data:
\( \Delta H_\mathrm{c}^\circ(\mathrm{CO}) = -283\ \mathrm{kJ\ mol^{-1}} \)
\( \Delta H_\mathrm{c}^\circ(\mathrm{CO_2}) = -394\ \mathrm{kJ\ mol^{-1}} \)
▶️ Answer / Explanation
Using combustion cycle:
\( \Delta H_\mathrm{r}^\circ = \Delta H_\mathrm{c}^\circ(\mathrm{reactants}) – \Delta H_\mathrm{c}^\circ(\mathrm{products}) \)
Reactants: CO combusts to \( \mathrm{CO_2} \): \( -283 \)
Products: \( \mathrm{CO_2} \): \( -394 \)
Overall:
\( \Delta H_\mathrm{r}^\circ = -283 – (-394) = +111\ \mathrm{kJ\ mol^{-1}} \)
Endothermic.
Example
Calculate the enthalpy change using bond energies for:
\( \mathrm{CH_3OH \rightarrow CO + 2H_2} \)
Bond energies (in \( \mathrm{kJ\ mol^{-1}} \)):
\( \mathrm{C-O = 358} \), \( \mathrm{C-H = 412} \), \( \mathrm{O-H = 463} \)
\( \mathrm{H-H = 436} \), \( \mathrm{C\equiv O = 1077} \)
▶️ Answer / Explanation
Bonds broken (reactant):
\( 3 \times \mathrm{C-H} = 3 \times 412 = 1236 \)
\( 1 \times \mathrm{C-O} = 358 \)
\( 1 \times \mathrm{O-H} = 463 \)
Total broken = \( 2057\ \mathrm{kJ\ mol^{-1}} \)
Bonds formed (products):
\( 1 \times \mathrm{C\equiv O} = 1077 \)
\( 2 \times \mathrm{H-H} = 2 \times 436 = 872 \)
Total formed = \( 1949\ \mathrm{kJ\ mol^{-1}} \)
Overall enthalpy change:
\( \Delta H_\mathrm{r} = 2057 – 1949 = +108\ \mathrm{kJ\ mol^{-1}} \)
Endothermic reaction.