State and explain Hess Law of constant Heat summation ?

Answer : Hess Law: ” The total heat change in a reaction is the same,whether the chemical reaction is the same,whether the chemical reaction takes place in a single step or in serveral steps “.

It is based on the Ist law of thermodynamics.

Explanation : Consider a reaction A—->D.

Suppose that D is formed from A in two different paths. Path ① and Path ②

Path ①:  ​ A → D, ΔH
Total heat change in path ①=ΔH

Path ②:  A → B : ΔH1​

 B→C :ΔH2

​ C→D: ΔH3​​

Total heat change in path ② is

ΔH₁​+ΔH₂+ΔH₃

Now ,from the Hess law,we have
ΔH = ΔH₁ +Δ₂ +ΔH3

3) Example:

Path ①:

           C(graphite) ​+O_2( g)​→CO_{2( g​)}  ;   ΔH=−393.5KJ mol^-1

Path ② :

          C(graphite) ​+1/2O_{2( g)}​→CO_(g)​;  ΔH_1​=−110.5KJ

          CO_(g)​+1/2O_{2( g)​}→CO_{2​( g)} ;    ΔH₂​=−283.02KJmol^-1

       Total heat change in path②  =ΔH_1​+ΔH₂

​       =(−110.5)+(−283.02)=−393.52KJ. 

Thus ΔH=ΔH1​+ΔH2​. Hence the Hess law is proved

                                   PROBLEMS IN Hess Law

1. The enthalpies of formation CO (g), CO₂ ​(g), N₂​O (g) and N_2​O_4​ (g) are − 110, − 393, 81 and 9.7 KJ/mol, respectively. Find the value of ΔH for the reaction.

N_2​O_4​ (g) + 3 CO → N₂​O (g) + 3 CO₂ ​(g).

Answer: ΔH^o = ΔH^o_(products) ​− ΔH^o_(reactants)​

ΔH ​= 3 ΔCO₂ ​+  ΔN_2​O ​− 3 ΔCO ​− Δ N_2​O_4​

​ΔH ​= 3 × (−393) + (81) − 3 × (−110) − (9.7)

ΔH ​= − (777.7) KJ​/mol

2.. Calculate the enthalpy change for the following reaction:

CH₄ ​(g) + 2 O₂ ​(g) ⟶ CO₂ ​(g) + 2 H_2​O (l).

Given that enthalpies of formation of CH₄, CO₂ and H_2​O are 74.8 kJmol⁻¹,− 393.5 kJ mol⁻¹, and − 286 kJmol⁻¹, respectively.

Answer: ΔH^o = ΔH^o_(products) ​− ΔH^o_(reactants)​

ΔH^o = [ΔH^o_(CO2) ​+ 2 X ΔH^o_(H2O)] − [ΔH^o_(CH4) + 2 X ΔH^o_(O2)​]

ΔH^o = [− 393.5 + 2 X (−286.2)] − [−74.8 + 2 X 0]

ΔH^o = − 393.5 – 572.4 + 74.8

ΔH^o = − 891.1 kJ/mol