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# Hess’s Law

## Statement of Hess’s Law

Hess’s law of constant heat summation states that the total enthalpy change in a particular reaction is constant regardless whether it occurs in one step or more.

## Explanation of Hess’s Law

According to Hess’s law, if A reacts to form the product B, it doesn’t matter how many steps involved to get the product, the total enthalpy change will be same. So the reaction can happen in one step where the enthalpy change is ΔH1. If it occurs in two steps, the enthalpy changes of these two steps are ΔHand ΔH3. If it is done in three steps, the enthalpy changes are ΔH, ΔHand ΔH6. So, according to the law,

ΔH= ΔH+ ΔH= ΔH+ ΔH+ ΔH

## Example

Formation of carbon dioxide from carbon and oxygen can occur in one step or two steps. In two steps it can be formed via the production of carbon monoxide.

In one step,

C(s) + O2(g) → CO2 ΔH1 = -393 kJ/mol ……………..(i)

In two steps,

step 1: C(s) + 1/2O2(g) → CO ΔH2 = -111 kJ/mol

Step 2: CO(g) + 1/2O2(g) → CO2 ΔH3 = -282 kJ/mol
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Overall change: C(s) + O2(g) → CO2 ΔH2+3 = -393 kJ/mol = ΔH1

This is same as equation (i) and the heat change also matches with that. So, ΔH= ΔH2+3. Thus it is proved that the Hess’s law is true.

## Applications

Using the Hess’s law the enthalpy of a particular reaction can be calculated. Sometime it is impossible to measure in any other way.

### Calculation of enthalpy of formation

The enthalpy change of formation can be calculated if we know the entalpy of formation of the reactants and products. As for example, the ethene gas and hydrogen chloride gas reacts to form chloroethane gas.

C2H4(g) + HCl(g) → C2H5Cl(g) ΔHf = ? kJ/mol ……………..(ii)

The enthalpy change of formation of this reaction need to be calculated. We can calculate that by knowing the heat of formation of ethene gas, hydrogen chloride gas and chloroethane gas from its elements.

2C(s) + 2H2(g) → C2H4(g) ΔHf1 = +52.2 kJ/mol……………….(iii)

1/2H2(g) + 1/2Cl2(g) → HCl(g) ΔHf2 = -92.3 kJ/mol…………(iv)

2C(s) + 2.5H2(g) + 1/2Cl2 → C2H5Cl(g) ΔHf3 = -109 kJ/mol………..(v)

So, adding (iii) and (iv), we get

2C(s) + 2.5H2(g) + 2Cl2(g) → C2H4(g) + HCl(g) ΔHf1+f2 = -40.1 kJ/mol…………(vi)

Then subtracting (vi) from (v), we get

C2H4(g) + HCl(g) → C2H5Cl(g) ΔHf = -68.9 kJ/mol ……………..(vii)

The equation (ii) and (vii) are same, so the enthalpy change of formation of chloroethane gas from ethene gas and hydrogen chloride gas is -68.9 kJ/mol.

### Calculation of enthalpy of formation from the enthapy of combustion

The enthalpy change of formation can be calculated if we know the entalpy of combustion of the reactants and products. As for example, the solid carbon and hydrogen gas reacts to form methane gas.

C(s) + 2H2(g) → CH4(g) ΔH1 = ? kJ/mol ……………..(viii)

The enthalpy change of formation of this reaction need to be calculated. We can calculated that by knowing the heat of combustion of carbon, hydrogen and methane.

C(s) + O2(g) → CO2(g) ΔHc1 = -393.7 kJ/mol……………….(ix)

H2(g) + 1/2O2(g) → H2O(l) ΔHc2 = -285.7 kJ/mol…………(x)

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ΔHc3 = -890.3 kJ/mol………..(xi)

So, doing (ix) + [2 X (x)], we get

C(s) + 2H2(g) + 2O2(g) → CO2(g) + 2H2O(l) ΔH2 = -965.1 kJ/mol…………(xii)

Then subtracting (xi) from (xii), we get

C(s) + 2H2(g) → CH4(g) ΔH1 = -74.8 kJ/mol…………(xiii)

The equation (viii) and (xiii) are same, so the enthalpy change of formation of methane gas from carbon and hydrogen is -74.8 kJ/mol.