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.
● 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.