Thermochemistry is the part of thermodynamics that studies the relationship between heat and chemical reactions. As we know thermodynamics is a very important topic from examination point of view. Hence as a part of it, thermochemistry also becomes important for exams. So if you want to be best in thermodynamics then you must study thermochemistry properly. So go ahead and prepare well. All the best !

**Q1.**For vaporization of water at 1 atmospheric pressure, the values of Î”H and Î”S are 40.63 kJ mol

^{−1}and 108.8 J K

^{−1}mol

^{−1}respectively. The temperature when Gibbs energy change (Î”G) for this transformation will be zero, is:

Solution:-

Î”G = Î”H − T.Î”S

Î”G = 0, ∴ Î”H = T.Î”S

T = Î”H / Î”S = 373.4 K

Î”G = Î”H − T.Î”S

Î”G = 0, ∴ Î”H = T.Î”S

T = Î”H / Î”S = 373.4 K

**Q2.**Î”n values in Î”H = Î”U + Î”nRT may have:

Solution:-

Î”n depends on stoichiometry of reaction.

Î”n depends on stoichiometry of reaction.

**Q3.**It is a general principle that if a system has the less energy then it is:

Solution:-

Lower is energy level of a system, more is its stability.

Lower is energy level of a system, more is its stability.

**Q4.**1 litre-atmosphere is equal to:

Solution:-

PV = 1 litre-atmosphere

= 10

= 101.3 J

PV = 1 litre-atmosphere

= 10

^{−3}m^{3}× 0.76 × 13.6 × 9.8 × 10^{3}Nm^{−2}= 101.3 J

**Q5.**A thermally isolated gaseous system can exchange energy with the surroundings. The mode of transference of energy can be:

Solution:-

Only work can be done by a thermally isolated system between it and surroundings.

Only work can be done by a thermally isolated system between it and surroundings.

**Q6.**Given that Î”H

_{r 298 K}= −54.07 kJ/mol and Î”S

^{°}

_{r 298 K}= 10 J/mol and R = 8.314 J K

^{−1}mol

^{−1}. The value of log

_{10}K for a reaction, A ⇌ B is:

Solution:-

Î”G° = Î”H° − TÎ”S°

= −54.07 − 298×10×10

= −57.05 kJ

Also, Î”G° = 2.303 RT log

Î”G° = Î”H° − TÎ”S°

= −54.07 − 298×10×10

^{−3}= −57.05 kJ

Also, Î”G° = 2.303 RT log

_{10}K**Q7.**If liquids A and B form an ideal solution, then:

Solution:-

Î”H

Î”H

_{mixing}= 0 for ideal solutions.**Q8.**The temperature at which the reaction,

Ag

_{2}O(s) → 2Ag(s) + 1/2O

_{2}(g)

is at equilibrium is ..... ; given, Î”H = 30.5 kJ mol

^{−1}and Î”S = 0.066 kJ mol

^{−1}K

^{−1}.

Solution:-

Î”G = Î”H − TÎ”S

Î”G = 0, at equilibrium; ∴ Î”H = TÎ”S

or 30.5 = T × 0.066

Î”G = Î”H − TÎ”S

Î”G = 0, at equilibrium; ∴ Î”H = TÎ”S

or 30.5 = T × 0.066

**Q9.**When 1g atom of carbon is converted into 1g molecule of CO

_{2}the heat liberated is same:

Solution:-

C + O

∴ Î”H = Î”U

C + O

_{2}(g) → CO_{2}(g), Î”n = 0;∴ Î”H = Î”U

**Q10.**For spontaneity of a cell, which is correct?

Solution:-

Î”G = Î”H − TÎ”S: Î”G = Î”E + pÎ”V − TÎ”S

For spontaneity Î”G = −ve

Î”G = Î”H − TÎ”S: Î”G = Î”E + pÎ”V − TÎ”S

For spontaneity Î”G = −ve