Test: JEE Previous Year Questions- Thermochemistry - JEE MCQ

ΔHReaction = ∑BE Reactant - ∑BE Product

= 4 x 414 + 615 + 435 – (6 x 414 + 347)

= 2706 – 2831

= –125 kJ

Hess's Law of Constant Heat Summation states that regardless of the multiple stages or intermediate steps of a reaction, the total enthalpy change for the reaction is the sum of all changes.

The enthalpy of a chemical process is independent of the path taken from the initial to the final state.

C + O₂  → CO₂
∆Hc = -393.5 kJ/mol    -i
CO + ½ O₂  → CO₂
∆Hc = -283 kJ/mol      -ii
On i-ii
We get 
 C + 1/2 O₂  → CO
∆Hf = -110.5 kJ/mol 

The correct answer is option D
Carbon is found in the solid-state. The state of the substance affects the enthalpy change.
Hence, the correct option is A

Oxidising power of chlorine in aqueous solution can be estimated by calculating the overall enthalpy change of reaction.
The reaction takes place with following steps:
1/2Cl2 —> Cl               ΔH1
Cl —> Cl−                    ΔH2
Cl− —> Cl− (aq)          ΔH3
ΔH will be the sum of the energy involved in these steps.
Hence, ΔH = ΔH1 +ΔH2 +ΔH3
ΔH1 = 240/2 = 120kJ/mol, ΔH2 = -349kJ/mol, ΔH3 = -381kJ/mol
Hence, ΔH = 120-349-381 = -610kJ/mol

The correct answer is option A
Given:- 
H₂​O_(l)​⟶H⁺_(aq.)​+OH⁻_(aq.)​ΔH=57.32kJ.....(1)
H_2​(g)​+
1/2O_2​(g)​⟶H₂​O_(l) ​ΔH=−286.20kJ.....(2)
Adding eqⁿ(1)&(2), we have
H_2​O_(l)​+H_2​(g)​+
1/2​O_2​(g)​⟶H⁺_(aq.)​+OH⁻_(aq.)​+H₂​O_(l)​;
ΔH=(57.32)+(−286.20)kJ
H_2​(g)​+
1/2​O_2​(g)​⟶H⁺_(aq.)​+OH⁻_(aq.)​;ΔH=−228.88kJ
Hence the value of enthalpy of formation of OH⁻ ion at 25℃ is −228.88kJ.

CH₃OH (l) + 3/2 O2 (g) → CO2 (g) + 2H2O(l)
∆G_r = ∆G_f(CO2,g) + 2∆G_f(H2O,l) - ∆G_f(CH3OH,l) - 3/2∆Gf(O2,g)
= -394.4 + 2(-237.2) - (-166.2) - 0
= -702.6 kJ
%age efficiency = 702.6/726 × 100 = 97%

The correct answer is option B
The change in the number of moles of gaseous species  Δn = 2−3 =−1
ΔH = ΔE+ΔnRT
−1366.5 kJ/mol = ΔE+[(−1)×8.314×10⁻³ kJ/mol/K×(27+273) K]
ΔE = −1364.0 kJ/mol
The value of internal energy change for the reaction is −1364.0 kJ/mol.
 

The correct answer is Option D.

From Hess’s Law;
ΔH_f + ΔHsub =  ΔHreaction
ΔH_f =  ΔHreaction - ΔHsub 
       =    - 111 kJ  -  54 kJ
       =    - 165 kJ
Thus the enthalpy of the formation (ΔHf) for N₂O₅(s) is -165 kJ