A-Level — chemistryMedium5 min read

Enthalpy Changes and Hess's Law

Master standard enthalpy changes, Hess's law, bond enthalpies, and calorimetry for A-Level Chemistry.

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# Enthalpy Changes and Hess's Law

Energetics is one of the most important areas of physical chemistry. Understanding enthalpy changes allows us to predict whether reactions will occur, calculate energy changes, and design industrial processes. This guide covers standard enthalpy definitions, Hess's law, calorimetry, and bond enthalpy calculations at A-Level.

1. Key Definitions

Enthalpy (HH) is the total energy content of a system at constant pressure. We measure the change in enthalpy (ΔH\Delta H).

Standard Conditions

Standard enthalpy changes are measured under standard conditions:

  • Temperature: 298 K298 \text{ K} (25°C25°\text{C})
  • Pressure: 100 kPa100 \text{ kPa} (1 bar)
  • Solutions: 1.00 mol dm−31.00 \text{ mol dm}^{-3}
  • Elements in their standard states (most stable form)

Symbol: ΔH⊖\Delta H^\ominus (the ⊖ means standard conditions)

Important Standard Enthalpy Changes

Name Symbol Definition
Enthalpy of combustion ΔcH⊖\Delta_c H^\ominus Enthalpy change when 1 mol of substance is completely burned in oxygen
Enthalpy of formation ΔfH⊖\Delta_f H^\ominus Enthalpy change when 1 mol of compound is formed from its elements in standard states
Enthalpy of neutralisation ΔneutH⊖\Delta_{neut} H^\ominus Enthalpy change when acid and base react to form 1 mol of water
Enthalpy of atomisation ΔatH⊖\Delta_{at} H^\ominus Enthalpy change when 1 mol of gaseous atoms is formed from the element in standard state
Mean bond enthalpy — Energy needed to break 1 mol of a particular bond in the gaseous phase, averaged over many compounds

2. Hess's Law

Hess's Law states:

The total enthalpy change for a reaction is independent of the route taken, provided the initial and final conditions are the same.

This is a consequence of the conservation of energy.

Using Hess's Law with Formation Enthalpies

ΔrH⊖=∑ΔfH⊖(products)−∑ΔfH⊖(reactants)\Delta_r H^\ominus = \sum \Delta_f H^\ominus (\text{products}) - \sum \Delta_f H^\ominus (\text{reactants})

Using Hess's Law with Combustion Enthalpies

ΔrH⊖=∑ΔcH⊖(reactants)−∑ΔcH⊖(products)\Delta_r H^\ominus = \sum \Delta_c H^\ominus (\text{reactants}) - \sum \Delta_c H^\ominus (\text{products})

(Note: the signs swap compared to formation!)

Worked Example: Using Formation Enthalpies

Problem

Question: Calculate ΔrH⊖\Delta_r H^\ominus for: CH4(g)+2O2(g)→CO2(g)+2H2O(l)\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)

Given: ΔfH⊖\Delta_f H^\ominus: CH₄ = −75 kJ/mol, CO₂ = −394 kJ/mol, H₂O = −286 kJ/mol, O₂ = 0

Solution

ΔrH=[(−394)+2(−286)]−[(−75)+2(0)]\Delta_r H = [(-394) + 2(-286)] - [(-75) + 2(0)] =(−394−572)−(−75)= (-394 - 572) - (-75) =−966+75=−891 kJ mol−1= -966 + 75 = -891 \text{ kJ mol}^{-1}

Worked Example: Using Combustion Enthalpies

Problem

Question: Calculate the enthalpy of formation of ethanol given:

  • ΔcH⊖\Delta_c H^\ominus: C = −394 kJ/mol, H₂ = −286 kJ/mol, C₂H₅OH = −1367 kJ/mol

Formation reaction: 2C(s)+3H2(g)+12O2(g)→C2H5OH(l)2\text{C}(s) + 3\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{C}_2\text{H}_5\text{OH}(l)

Solution

ΔfH=∑ΔcH(reactants)−∑ΔcH(products)\Delta_f H = \sum \Delta_c H(\text{reactants}) - \sum \Delta_c H(\text{products}) =[2(−394)+3(−286)]−[−1367]= [2(-394) + 3(-286)] - [-1367] =[−788−858]+1367= [-788 - 858] + 1367 =−1646+1367=−279 kJ mol−1= -1646 + 1367 = -279 \text{ kJ mol}^{-1}

Worked Example: Bond Enthalpy Calculation

Problem

Question: Calculate ΔH\Delta H for CH4+2O2→CO2+2H2O\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} using mean bond enthalpies.

Bond enthalpies: C−H = 413, O=O = 498, C=O = 805, O−H = 464 kJ/mol

Solution

Bonds broken: 4(C−H) + 2(O=O) = 4(413) + 2(498) = 1652 + 996 = 2648 kJ

Bonds made: 2(C=O) + 4(O−H) = 2(805) + 4(464) = 1610 + 1856 = 3466 kJ

ΔH=2648−3466=−818 kJ mol−1\Delta H = 2648 - 3466 = -818 \text{ kJ mol}^{-1}

Note: Bond enthalpy calculations give approximate values because they use mean (average) values. Actual bond energies vary between different molecules.

4. Calorimetry

Calorimetry measures enthalpy changes experimentally.

The Key Equation

q=mcΔTq = mc\Delta T

where:

  • qq = energy transferred (J)
  • mm = mass of solution (g) — usually assume density = 1 g cm−31 \text{ g cm}^{-3}
  • cc = specific heat capacity (4.18 J g−1K−14.18 \text{ J g}^{-1} \text{K}^{-1} for water)
  • ΔT\Delta T = temperature change (K or °C)

Then: ΔH=−qn\Delta H = \frac{-q}{n} where nn = moles of the substance of interest.

Example: Calorimetry

Question: 50 cm³ of 1.0 mol/dm³ HCl is mixed with 50 cm³ of 1.0 mol/dm³ NaOH. Temperature rises from 22.0°C to 28.8°C. Calculate ΔneutH\Delta_{neut} H.

Solution: q=mcΔT=100×4.18×6.8=2842 J=2.84 kJq = mc\Delta T = 100 \times 4.18 \times 6.8 = 2842 \text{ J} = 2.84 \text{ kJ}

Moles of water formed = 0.050 mol (from 50 cm³ × 1.0 mol/dm³)

ΔH=−28420.050=−56840 J mol−1=−56.8 kJ mol−1\Delta H = \frac{-2842}{0.050} = -56840 \text{ J mol}^{-1} = -56.8 \text{ kJ mol}^{-1}

5. Sources of Error in Calorimetry

  • Heat loss to the surroundings (main source)
  • Incomplete combustion
  • Assumptions about specific heat capacity and density
  • Heat absorbed by the calorimeter itself
  • Temperature measurement inaccuracy

Improvements: use a bomb calorimeter (sealed, insulated); use a lid and insulation; extrapolate cooling curves.

6. Practice Questions

    1. Define standard enthalpy of formation and standard enthalpy of combustion.
    1. Use Hess's law and the following data to calculate ΔfH\Delta_f H of propane: ΔcH\Delta_c H: C(s) = −394, H₂(g) = −286, C₃H₈(g) = −2220 kJ/mol
    1. Calculate ΔH\Delta H for N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 using bond enthalpies: N≡N = 945, H−H = 436, N−H = 391 kJ/mol
    1. 100 cm³ of 0.50 mol/dm³ CuSO₄ reacts with excess zinc. Temperature rises by 12.5°C. Calculate the enthalpy change per mole of CuSO₄.
    1. Explain why bond enthalpy calculations give different values from those using formation data.

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7. Exam Tips

  • Pay attention to the direction of Hess cycles — products minus reactants for formation, reactants minus products for combustion
  • In bond enthalpy calculations: break ALL bonds in reactants, make ALL bonds in products
  • For calorimetry: q=mcΔTq = mc\Delta T then divide by moles and add a negative sign (if exothermic)
  • Always state the sign of ΔH\Delta H — positive or negative matters!
  • Elements in standard states have ΔfH=0\Delta_f H = 0 by definition

Summary

  • Hess's Law: total ΔH\Delta H is route-independent
  • Using formation: ΔH=∑ΔfH(products)−∑ΔfH(reactants)\Delta H = \sum \Delta_f H(\text{products}) - \sum \Delta_f H(\text{reactants})
  • Using combustion: ΔH=∑ΔcH(reactants)−∑ΔcH(products)\Delta H = \sum \Delta_c H(\text{reactants}) - \sum \Delta_c H(\text{products})
  • Bond enthalpies: ΔH=bonds broken−bonds made\Delta H = \text{bonds broken} - \text{bonds made}
  • Calorimetry: q=mcΔTq = mc\Delta T, then ΔH=−q/n\Delta H = -q/n
Tags:
#a-level#chemistry#energetics#enthalpy#hess-law

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