Hess's Law and enthalpy cycles

​​In a nutshell

Understanding Hess's law and how this can be applied to enthalpy cycles allows you to calculate enthalpy changes indirectly. By interpreting different enthalpy values, you can learn to produce enthalpy cycles and determine unknown enthalpy changes. 

Enthalpy cycles 

Hess's law states that regardless of the route you take, the total enthalpy change will be equal. 

This law applies to enthalpy cycles which show alternative routes for a reaction to reach the same product. Any route will result in the same total enthalpy change. 

Standard enthalpies of formation

Enthalpy cycles are useful to determine unknown enthalpy changes if all $$\varDelta _f H^\theta$$​​ are known for compounds.​

Note: Elements always have an $$\varDelta _f H^\theta$$ of zero. 

​$$\varDelta _r H^\theta$$ is the standard enthalpy change of reaction where compounds have their molar quantities in the equation under standard conditions.

$$\varDelta _f H^\theta$$ is the standard enthalpy change of formation where one mole of a compound is formed from its individual elements in standard states under standard conditions.

PROCEDURE 

Example

​$$\varDelta _fH^\theta\ [H_2O(l)]=-84\ kJ\ mol^{-1}$$

$$\varDelta _f H^\theta [ CO_2(g) ]=-281\ kJ\ mol^{-1}$$​​​

What is $$\varDelta _r H^\theta$$? 

Following Hess's law, route 1 and route 2 will have an equal enthalpy change. The arrows for $$\varDelta _fH ^\theta$$ point towards the main reaction as the reactants and products at the top are formed from the elements at the bottom. $$\varDelta _fH ^\theta$$​ is presented for all the compounds present in the chemical reaction. All elements ($$H_2, O_2, C$$​) have an $$\varDelta _fH^\theta$$​ of zero.

If route 1 = route 2:

$$\varDelta _f H^\theta_{reactants} +\varDelta _r H^\theta=\varDelta _f H^\theta_{products}$$

This can be rearranged to:

$$\varDelta _r H^\theta=\varDelta _f H^\theta_{products}-\varDelta _f H^\theta_{reactants}$$​​

​

Now you can input the given values into this equation:

$$\varDelta_r H^\theta=[(CO_2)+ (2\times H_2)]-[(2\times H_2O)+ (C)]$$​​

$$\varDelta _r H^\theta=[-281+(2\times0)]-[(2\times-84)+0]$$

$$\varDelta _r H^\theta=-113\ kJ\ mol^{-1}$$

​$$\varDelta _r H^\theta$$ for this reaction is $$\underline{-113\ kJ\ mol^{-1}}$$. 

​

Note: The first zero represents the element $$H_2$$ in the products and second zero represents the element $$C$$ in the reactants.

Standard enthalpies of combustion

​Similarly, enthalpy cycles can use $$\varDelta_cH^\theta$$ to determine unknown enthalpy changes. All reactants and products other than oxygen will have a $$\varDelta_cH^\theta$$ value. 

Note: Combustion products do not have a $$\varDelta_cH^\theta$$ value. 

$$\varDelta_cH^\theta$$ is the standard enthalpy change of combustion where one mole of a substance is completely combusted in oxygen under standard conditions.

​ PROCEDURE

Example 

​$$\varDelta_cH^\theta[C(s)]=-389\ kJ\ mol^{-1}$$​​

​$$\varDelta_cH^\theta[H_2{(s)}]=-315\ kJ\ mol^{-1}$$​​

​$$\varDelta_cH^\theta[C_4H_9OH(l)]=-2479\ kJ\ mol^{-1}$$​​

What is $$\varDelta_fH^\theta$$?

As route 1 and 2 have an equal enthalpy change, we can determine $$\varDelta_fH^\theta$$ from the $$\varDelta_cH^\theta$$ values provided. The arrows for $$\varDelta_cH^\theta$$ point from the main reaction to the bottom compounds as the reactants and products at the top are combusted to produce the same combustion products shown at the bottom. The combustion products and oxygen have $$\varDelta_cH^\theta$$ of zero.

If route 1 = route 2:

$$\varDelta_fH^\theta+\varDelta_cH^\theta\ _{(products)}=\varDelta_c H^\theta\ _{(reactants)}$$​​

This can be rearranged to:

$$\varDelta_fH^\theta=\varDelta_c H^\theta\ _{(reactants)}-\varDelta_cH^\theta\ _{(products)}$$​​

Input values for the corresponding compounds:

$$\varDelta_fH^\theta=[(4\times C)+(5\times H_2)+(\frac{1}{2}\times O_2)]-[C_4H_9OH]$$

$$\varDelta_fH^\theta=[(4\times -389)+(5\times-315)+(\frac{1}{2}\times0)]-[-2479]$$​​

​$$\varDelta_fH^\theta\ =-652\ kJ\ mol^{-1}$$

​$$\varDelta_fH^\theta$$ for this reaction is $$\underline{-652\ kJ\ mol^{-1}}$$.

Note: $$\varDelta_cH^\theta$$ can be carried out using individual elements or hydrocarbons as the reactants.

Forming enthalpy cycles 

Enthalpy cycles (or Hess cycles) can be formed when you cannot directly measure enthalpy changes such as thermal decomposition reactions. These are endothermic so decrease in temperature, but must be heated to begin the reaction. Calorimetry experiments can be carried out to measure the total enthalpy changes of an alternative route that is equal to the route you cannot measure.

PROCEDURE   

Example

Draw an enthalpy cycle for the thermal decomposition of carbonic acid, $$H_2CO_3$$.  

Write out the equation for the thermal decomposition of carbonic acid:

$$H_2CO_3(l)\rightarrow CO_2(g)+ H_2O(l)$$​​

​

This will form the top of the enthalpy cycle which is $$\varDelta _r H^\theta$$. A calorimetry experiment can be carried out to measure $$\varDelta _f H^\theta$$ of the reactants and products which would give you:

$$H_2(g)+C(s)+1\frac{1}{2}O_2(g)$$​​

​

This will form the bottom of the enthalpy cycle. The arrow towards the reactants of the thermal decomposition will be $$\varDelta _fH^\theta\ _{(reactants)}$$ and the arrow towards the products will be $$\varDelta _fH^\theta\ _{(products)}$$​.

Now you can form the enthalpy cycle which will look like this:

Both of the bottom arrows are pointing towards the main reaction as the elements at the bottom are forming the reactants and products of the main reaction. The two routes can be distinguished by the different colour arrows.

​

$$\varDelta_r H^\theta$$ for the thermal decomposition of carbonic acid can be calculated by using the general formula route 1 = route 2:

​$$\varDelta_f H^\theta_{(products)}+\varDelta_r H^\theta=\varDelta_f H^\theta_{(reactants)}$$​​

$$\varDelta_r H^\theta=\varDelta_f H^\theta_{(products)}-\varDelta_f H^\theta_{(reactants)}$$​

These numbers can be determined through calorimetry experiments.