Test: Enthalpy of Reaction & Formation - NEET MCQ

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

∆H= ∆H_CO+∆H_H2O-∆H_CO2 (as ΔH for H₂ is 0)
= -110.5-241.8+393.5 = +41.2 kJ mol^-1

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

For this, we will add combustion reactions of both CO and H₂ as both combined will give us energy.
Δ_rH = -283 kJ+ (-242 kJ) = -525 kJ

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

For ∆H formation we have to look for following:
1. formation of "1 mole" of substance.
2. from "stable" constituting elements.

In first option 1 mole of Co₂ should be formed from C(graphite) for enthalpy of formation to be defined.
In third option 2 moles of NH₃ is formed instead of 1 mole. So incorrect.
In fourth option Co₂ is not formed from constituent elements I.e C(gr) and O2(g) so it is not enthalpy of formation.

A/Q molar ratio of C₂H₄ and CH₃CHO is 8:1
So from eqn 1
∆H = 8 45.54
= 364.32 kJ
Similarly from eqn 
∆H = 68.91 1 
= 68.91 kJ
Total ∆H = 364.32+68.91
Therefore, enthalpy change per mole of C₂H₅OH used 364.32+68.91/9
= 48.14 kJ

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.

Enthalpy of formation means that the reactants are in their pure state and one mole of product is formed.
For enthalpy of formation of CO₂ we need eqn i/2 + eqn ii.
C(graphite) + O₂(g) →   CO₂  ∆H = -64 kcal mol-1
For enthalpy of formation of CO, eqn i/2
C(graphite) + O₂(g) →  CO   ∆H = -26 kcal mol^-1

The correct answer is option C
From, Kirchhoff's equation
△H₂​ (1000 K) = △H1​(300 K) + △C_p​(1000−300)
Here, △H2​(1000 K) = −123.77 kJ mol⁻¹
△H1​(300 K)=?
△C_p​ = 2C_p​(NH_3​) − [C_p​(N₂​) + 3C_p​(H₂​)]
=−6R
=−6 × 8.314 × 10⁻³kJ
∴−123.77 = △H1​(300 K) + 6 × 8.314 × 10⁻³× 700
or △H^1​(300 K) = −88.85 kJ
For two moles of NH_3​
∴△H_f​ (NH_3​) = △H1​(300K)/2​
=−88.85/​2
=−44.42 kJ mol⁻¹

Mass of O₂=32 gm ∆H/gm = -483.636/32 = -15.11
Mass of O₃=48 gm ∆H/gm = -868.2/48 = -18.08
Mass of H₂O₂=34 gm, ∆H/gm = -347.33/34 = -10.21
We can say that II has highest amount of energy per gram of oxidising agent.

From I, total mass of reactant, 36 gm
So, ∆H/mass = -483.636/36 = -13.43 kJ gm^-1
For II, total mass of reactant, 54 gm
So, ∆H/mass = -868.2/54 = -16.07 kJ gm^-1
FOr III, total mass = 36 gm,  
So, ∆H/mass = -347.33/36 = -9.64 kJ gm^-1
Therefore, heat released per unit mass is maximum for II.

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

The standard molar enthalpy of formation of elements in its standard state is taken as zero. Since the standard state of H₂ and Cl₂ are gas, their ΔfHo = 0.