Thermodynamics is a very important branch of both chemistry and physics. It deals with the study of energy, the conversion of energy between different forms and the ability of energy to do work. Thermodynamics is an important chapter for JEE Mains, JEE Advance NEET and many others exams. As this chapter is present in both chemistry and physics and there is only a minor difference between them hence it becomes more important topic. It is not a tough topic, it only requires practice. This topic has been given good weightage in all engineering exams. So don't skip this and prepare well. All the best !.

**Q1.**The amount of heat measured for a reaction in a bomb calorimeter is

Solution:-

Bomb calorimeter measures q

Bomb calorimeter measures q

_{v}, which is equal to Î”E.**Q2.**The standard heat of formation of NO

_{2}(g) and N

_{2}O

_{4}(g) are 8.0 and 2.0 kcal/mol respectively. The heat of dimerization of NO

_{2}in kcal is:

Solution:-

**Q3.**Which reaction is endothermic in nature ?

Solution:-

Decomposition of CaCO

Decomposition of CaCO

_{3}is made on heating.**Q5.**Joule-Thomson expansion is

Solution:-

Isoenthalpic

Isoenthalpic

**Q6.**Heat of formation of H

_{2}O(g) at 1 atm and 25°C is −243 kJ. Î”U for the reaction,

H

_{2}(g) + 1/2 O

_{2}(g) → H

_{2}O(g) at 25°C is :

Solution:-

Î”H = Î”U + Î”nRT

Î”n = − 1/2

∴ −43 = Î”U + (−1/2) × 8.314 × 298 × 10

∴ Î”U = −241.76 kJ

Î”H = Î”U + Î”nRT

Î”n = − 1/2

∴ −43 = Î”U + (−1/2) × 8.314 × 298 × 10

^{−3}∴ Î”U = −241.76 kJ

**Q7.**Diborane is a potential rocket fuel which undergoes combustion according to equation :

B

_{2}H

_{6}(g) + 3O

_{2}(g) → B

_{2}O

_{3}(g) + 3H

_{2}O(g)

Calculate the enthalpy change for the combustion of diborane. Given :

Solution:-

For the equation

B

eq.(i) + 3(ii) + 3(iii) − (iv)

Î”H = −1273 + 3(−286) + 3(44) − 36

= −2035 kJ/mol

For the equation

B

_{2}H_{6}(g) + 3O_{2}(g) → B_{2}O_{3}(g) + 3H_{2}Oeq.(i) + 3(ii) + 3(iii) − (iv)

Î”H = −1273 + 3(−286) + 3(44) − 36

= −2035 kJ/mol

**Q8.**In a closed insulated container a liquid is stirred with a paddle to increase the temperature, which of the following is true ?

Solution:-

For insulated container, q = 0.

For insulated container, q = 0.

**Q9.**C(s) + O

_{2}(g) → CO

_{2}(g), Î”H = −393.5 kJ and

CO(g) + 1/2 O

_{2}(g) → CO

_{2}(g), Î”H = −283.5 kJ

then the heat of formation of CO is :

Solution:-

C(s) + O

CO(g) + 1/2 O

On subtracting, C(s) + 1/2 O

C(s) + O

_{2}(g) → CO_{2}(g); Î”H = −393.5 kJCO(g) + 1/2 O

_{2}(g) → CO_{2}(g); Î”H = −283.5 kJOn subtracting, C(s) + 1/2 O

_{2}(g) → CO(g); Î”H = −110.0 kJ**Q10.**Equal volume of 1M HCl and 1M H

_{2}SO

_{4}are neutralised by dil. NaOH solution and x and y kcal of heat are liberated respectively. Which of the following is true ?

Solution:-

1M H

1M HCl = 1 eq. HCl

Thus, for equal volume of two acids to be neutralized separately with NaOH, heat evolved will be twice in case of H

1M H

_{2}SO_{4}= 2 eq. H_{2}SO_{4}1M HCl = 1 eq. HCl

Thus, for equal volume of two acids to be neutralized separately with NaOH, heat evolved will be twice in case of H

_{2}SO_{4}to that of HCl.