Test: Enthalpy of Neutralisation & Solution - NEET MCQ

Enthalpy of reaction = ∑enthalpy of products - ∑enthalpy of reactants
Therefore
∆H_reaction = -167.44 - (- 92.30) (∆H_formation of H+ is 0)
= -75.14 kilo J / mole

∆H_{neutralisation} =  ∆H_H⁺ _{+ OH}- + ∆_rHionisation 
-5.1 = -57.3 + ∆H_ionisation
∆H_ionisation = 52.2 kJ mol^-1

½ H₂(g) + ½ Cl₂(g) → HCl(g) ∆_fH° = -22.10 kcal mol^-1  -----(I)
HCl(g) + H₂O(l) → H⁺(aq) + Cl^-(aq)     ∆_solutionH° = -17.9 kcal mol^-1 -----(II)
Let us add (I) and (II), we get
½ H₂(g) + ½ Cl₂(g) → H⁺(aq) + Cl-(aq) ∆_rH° = -40.0 kcal mol^-1
∆_rH° = ∑∆H_product  -  ∑∆H_reactant = ∆_fH_Cl- (since enthalpy of formation of H, H₂ and Cl₂ are 0)
∆_fH_Cl- = -40.0

Enthalpy of neutralisation = ∆H_{H+ + OH- → H2O +} ∆H_ionisation
-24.0 = -27.4 + ∆H_ionisation   (since we have 2 moles of H⁺ and OH^-, so enthalpy  will also double)
∆H_ionisation = 3.4 kcal mol-1

CH₃COOH → CH₃COO^- + H⁺
∆H = 0.3 kcal mol^-1
H⁺ + OH^- → H₂O
∆H = -13.7 kcal mol-1
Equation needed
2CCH₃COH + Ca(OH)₂ → 2CH₃COO + 2H₂O
∆H = 2×EQN(I) + 2×EQN(II) 
∆H = (0.3)2 + (-13.7)2 = -26.8 kcal mol^-1

For H⁺(aq) + OH^-(aq) → H₂O(l), we need to have -(I)+(II).
On doing the, we have ∆H = -14.22 + 0.66 = -13.56 kcal

We can say that enthalpy of neutralisation of H₂C₂O4 means that we have 2 moles of H⁺. So, for weak acid, enthalpy of neutralisation = heat released in neutralisation of H⁺ and OH^- + enthalpy of ionisation.
∆H_{neutralisation} = 2 (-13.6) + 3.2 = -24.2 kcal mol^-1

H₂S(aq) + OH^-(aq) → HS-(aq) + H₂O(l)
(l) ∆H_r° = -33.7 kJ mol^-1
We also have, H+(aq) + OH-(aq)  →  H2O(l)
(ll) ∆H_r° = -57.3 kJ mol^-1
For the reaction, 
we need (I) -(II)
So the Δ_rH° be = -33.7 - (-57.3) = 23.6 kJ mol^-1

Li+ + H₂O  →  Li⁺(aq.)
∆H₁ = -499 kJ mol^-1 ---------- (I)
Cl^- + H₂O →Cl^-(aq)
∆H₂ = -382 kJ mol^-1 ---------- (II)          
Li⁺ + Cl^- → LiCl(s)
∆H₃ = -840 kJ mol^-1---------- (III)
For r the heat of solution, we need the reaction,
LiCl(s) → Li⁺(aq.) + Cl^-(aq)
This can be obtained by (I) + (II) - (III)
So the ∆H value be = -499 - 382 -(-840) = -41 kJ mol^-1.

For reaction of 1 mole of H⁺ and 1 mole of OH^-, -13.6 kcal energy is released.
100 mL of 1 M Ca(OH)₂ = moles = molarity volume = 0.1× 1 = 0.1
Moles of OH^- = 0.2
 100 mL of 1 M H₂SO₄=  moles = molarity volume = 0.1× 1 = 0.1
Moles of H⁺ = 0.2
SO 0.2 moles of H⁺ and OH^- reacts. 
Energy released, 0.2 ×13.6 =2.74 kcal.

Since both reactions are acid base neutralization having strong acid-strong base and in 1:1 ratio of H⁺ and OH^- ,so will have same enthalpy of reaction equal to-13.6 kcal.

On (I) - (II), we have
HF(aq)  →  H⁺(aq) + F^-(aq)
∆H = -16.4-(-13.7) = -2.7 kcal
So, the ionisation of HF is an exothermic reaction. And so the statement I is true.
The reason behind HF being exothermic as the hydration energy of H⁺ and F^- exceeds the lettuce enthalpy of HF and so overall the reaction becomes exothermic.

Milliequivalents of the acid taken = 4×2 =8. 
neutralization of 8 meq liberated 64 cal heat. 
So the heat of neutralization is 64/8 = 8cal/meq.

For reaction of 1 mole of H⁺ and one mole of OH^-, -13.6 kcal energy is released. As we have energy released = -54.8, 
Therefore -13.6×n_H+ = -54.8
n_H+ = 4
So, the basicity of acid given here is 4.