Test: Enthalpy Change ∆rH of a Reaction – Reaction Enthalpy (June 14) - NEET MCQ

∆rH° =  ∆rE° + ∆ngRT
∆rH° - ∆rE° = ∆ngRT
= (12-15)× 2× 298
= -1.788 kcal

Δ_rH=Δ_rH_(product)-ΔrH_(reactant)
=3×-393+81+3×110-9.7
=-777.7kJ
ΔH = ΔU+Δn_gRT
AS Δn_g = 0, ΔH = ΔU
So, ΔU = -777.7 kJ

Let's number our eqns-
CH₃OH + 3/2 O₂ → CO₂ + 2H₂O ...(1)
ΔH₁ = –726 kJ/mol
C + O₂ → CO₂ ...(2)
ΔH₂ = –393 kJ/mol
H₂ + 1/2 O₂ → H2O ...(3)
ΔH₃ = –286 kJ/mol
Eqn(2) + 2×eqn(3) - eqn(1) :-
C + O₂ + 2H₂ + O₂ + CO₂ + 2H₂0→ CO₂ + 2H₂0 + CH₃OH + 3/2 O₂
C + 1/2 O₂ + 2H₂ → CH₃OH
Thus enthalpy of formation-
∆Hf(CH₃OH) = ∆H₂ + 2∆H₃ - ∆H₁
∆H_f(CH₃OH) = -393 + 2(-286) + 726
∆Hf(CH3OH) = -239 kJ/mol

ΔH=ΔU+Δ(PV)
ΔH=30+(P₂V₂−P₁V₁)
ΔH=30+(20−6)
ΔH=30+14=44

Reaction number 1 is less stable because heat is not released thus the product side of reaction gains energy which get stored in the product and decreases its stability
While in reaction number 2, the reaction formed is more stable because energy is released and so the product is in a more stable state.
Now there is no such endothermic or exothermic compound , the reaction is exothermic or endothermic and not the product.

C(diamond) + O₂(g) → CO₂(g) -----(i)     ∆_rH°  = -91 kcal mol^-1
C(graphite) + O₂(g) → CO₂(g) -----(ii)     ∆_rH°  = -94 kcal mol^-1
For 36 gm of C(diamond), we have 3 moles of diamond, 
On (i)-(ii), we get → C(graphite)  ∆rH°  =-91-( -94 ) kcal mol^-1 = 3 kcal mol^-1
So for 3 moles, we have 3×3 = 9 kcal 
C(diamond) 

N₂(g) + O₂(g) → 2NO ----- (i) ∆_rH° = 43 kcal
2NO(g) + O₂(g) → 2NO₂(g) -----(ii)     ∆rH° = -35 kcal
Enthalpy of formation of NO₂ means we need to have 1 mole of NO₂ from its constituting elements. 
We will have (i)/2 + (ii)/2
½ N₂(g) + O₂(g) → NO₂(g) ∆_fH° = 4 kcal