Chemistry Thermochemical Equations and Hess's Law

Topics Covered :

● Thermochemical Equations
● Hess's Law of Constant Heat Summation

Thermochemical equations :

`=>` A balanced chemical equation together with the value of it's `Δ_r H` is called a `color{red}("thermochemical equation")`.

● We specify the physical state (alongwith allotropic state) of the substance in an equation.

● For example :

`color{purple}(C_2H_5 OH (l) +3O_2(g)→ 2CO_2(g) +3H_2O(l) ; Delta_r H^⊖ = -1367 kJ mol^(-1))`

● The above equation describes the combustion of liquid ethanol at constant temperature and pressure.

● The negative sign of enthalpy change indicates that this is an exothermic reaction.

`=>` It would be necessary to remember the following conventions regarding thermochemical equations.

(i) The coefficients in a balanced thermochemical equation refer to the number of moles (never molecules) of reactants and products involved in the reaction.

(ii) The numerical value of `color{purple}(Δ_rH^⊖)` refers to the number of moles of substances specified by an equation. Standard enthalpy change `color{purple}(Δ_rH^⊖)` will have units as `kJ mol^(–1)`.

`color{red}("Example ")` Let us consider the calculation of heat of reaction for the following reaction :

`color{purple}(Fe_2O_3 (s) +3H_2(g)→ 2Fe (s) +3H_2O (l))`

From the Table (6.2) of standard enthalpy of formation `color{purple}(Δ_f H^⊖)`, we find :

`color{purple}(Delta_f H^⊖ (H_2O , l) = -285.83 kJ mol^(-1))`

`color{purple}(Delta_f H^⊖ (Fe_2O_3 , s) = -824.2 kJ mol^(-1))`

Also `color{purple}(Delta_fH^⊖ ( Fe, s) = 0)` and `color{purple}(Delta_f H^⊖ ( H_2 , g) = 0)` as per convention

Then `color{purple}(Delta_r H_1^⊖ = 3 (-285.83 kJ mol^(-1) ) -1 ( -824.2kJ mol^(-1)))`

` color{purple}(= (-857.5+824.2) kJ mol^(-1))`

`color{purple}( = -33.3 kJ mol^(-1))`

`color{red}("Note ")` The coefficients used in these calculations are pure numbers, which are equal to the respective stoichiometric coefficients.

● The unit for `color{purple}(Δ_rH^⊖)` is `kJ mol^(–1)`, which means per mole of reaction.

● Once we balance the chemical equation in a particular way, as above, this defines the mole of reaction. If we had balanced the equation differently, for example,

`color{purple}(1/2 Fe_2O_3 (s) +3/2H_2(g) → Fe (s) +3/2 H_2O (l))`

then this amount of reaction would be one mole of reaction and `color{purple}(Δ_rH^⊖)` would be

`color{purple}(Delta_r H_2^⊖ = 3/2 (-285.83 kJ mol^(-1)) -1/2 (-824.2 kJ mol^(-1)))`

`color{purple}(= (-428.7+412.1) kJ mol^(-1))`

`color{purple}(= -16.6 kJ mol^(-1) = 1/2 Delta_r H_1^⊖)`

● It shows that enthalpy is an extensive quantity.

(iii) When a chemical equation is reversed, the value of `color{purple}(Δ_rH^⊖)` is reversed in sign.

`color{red}("Example ")`

`color{purple}(N_2 (g) +3H_2 (g ) → 2NH_3(g) ; Delta_r H^⊖ = -91.8 kJ mol^(-1) )`

`color{purple}(2NH_3 (g) → N_2(g) +3H_2(g) ; Delta_r H^⊖ = +91.8 kJ mol^(-1))`

Hess’s Law of Constant Heat Summation :

`=>` We know that enthalpy is a state function, therefore the change in enthalpy is independent of the path between initial state (reactants) and final state (products).

● In other words, enthalpy change for a reaction is the same whether it occurs in one step or in a series of steps.

● This may be stated as follows in the form of Hess’s Law.

`color{green}("If a reaction takes place in several steps then its standard reaction enthalpy")` `color{green}("is the sum of the standard enthalpies of the intermediate reactions")` `color{green}("into which the overall reaction may be divided at the same temperature")`.

`=>` Let us understand the importance of this law with the help of an example.

● Consider the enthalpy change for the reaction

`color{purple}(C_text(graphite,s) +1/2 O_2 (g) → CO (g) ; Delta_f H^⊖ = ?)`

● Although `color{purple}(CO(g))` is the major product, some `color{purple}(CO_2)` gas is always produced in this reaction.

● Therefore, we cannot measure enthalpy change for the above reaction directly.

● However, if we can find some other reactions involving related species, it is possible to calculate the enthalpy change for the above reaction.

● Let us consider the following reactions :

`color{purple}(C_text(graphite,s) +O_2(g) → CO_2(g) ; Delta_r H^⊖ = -393.5 kJ mol^(-1))` .................... (i)

`color{purple}(CO(g)+1/2O_2( g) → CO_2(g) ; Delta_r H^(⊖) = -283.0 kJ mol^(-1))` ...(ii)

● We can combine the above two reactions in such a way so as to obtain the desired reaction.

● To get one mole of `color{purple}(CO(g))` on the right, we reverse equation (ii). In this, heat is absorbed instead of being released, so we change sign of `color{purple}(Δ_rH^⊖)` value

`color{purple}(CO_2 (g) → CO(g) +1/2O_2(g)) ; `

`color{purple}(Delta_r H^⊖ = +283)` ............(iii)

Adding equation (i) and (iii), we get the desired equation,

`color{purple}(C_text(graphite,s) +1/2 O_2 (g) → CO (g)) ; `

for which `color{purple}(= Δ_r H^⊖ = (-393.5+283.0))`
`color{purple}(= – 110.5 kJ mol^(–1))`

`=>` In general, if enthalpy of an overall reaction `color{purple}(A→B)` along one route is `color{purple}(Δ_r H)` and `color{purple}(Δ_rH_1, Δ_rH_2, Δ_rH_3)`..... representing enthalpies of reactions leading to same product, `B` along another route, then we have

`color{purple}(Δ_rH = Δ_rH_1 + Δ_rH_2 + Δ_rH_3)` ...(6.16)

● It can be represented as shown in fig.