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. Given that C(g) + 4H(g) → CH4(g); ΔH = −166kJ. The bond energy C − H will be:
Solution:-
Bond-energy of C − H bond = −166/4
= −41.5 kJ/mol
Bond-energy of C − H bond = −166/4
= −41.5 kJ/mol
Q2. If S° for H2, Cl2 and HCl are 0.13, 0.22 and 0.19 kJ K−1 mol−1 respectively. The total change in standard entropy for the reaction, H2 + Cl2 → 2HCl is:
Solution:-
ΔS = SP − SR
= (2×0.19)−0.13−0.22 = 0.03 kJ K−1 mol−1
= 30 J K−1 mol−1
ΔS = SP − SR
= (2×0.19)−0.13−0.22 = 0.03 kJ K−1 mol−1
= 30 J K−1 mol−1
Q3. Hess law is applicable for the determination of heat of
Solution:-
All of these
All of these
Q4. The Kirchhoff’s equation gives the effect of ...... on heat of reaction.
Solution:-
Kirchhoff’s equation is : ΔH2 − ΔH1 = ΔCp(T2 − T1)
Kirchhoff’s equation is : ΔH2 − ΔH1 = ΔCp(T2 − T1)
Q5. The enthalpy change for the following reaction, NaOH(aq) + HCl(aq) → NaCl(aq) + H2(l) is −57 kJ. Predict the value of the enthalpy change in the following reaction : Ba(OH)2 + H2SO4(aq) → BaSO4(s) 2H2O(l)
Solution:-
Two equivalent of each are used.
Two equivalent of each are used.
Q6. From the reaction Pwhite → Pred; ΔH = −18.4 kJ it is following that :
Solution:-
Energy of red P is lesser than white P and thus, red P is more stable.
Energy of red P is lesser than white P and thus, red P is more stable.
Q7. Heat evolved in the reaction, H2 + Cl2 → 2HCl is 182 kJ. Bond energies of H − H and Cl − Cl are 430 and 242 kJ/mol respectively. The H − Cl bond energy is :
Solution:-
ΔH = −2×eH−Cl + eH−H + eCl−Cl
∴ 182 = −2×a + 430 + 242
∴ a = 245 kJ/mol
ΔH = −2×eH−Cl + eH−H + eCl−Cl
∴ 182 = −2×a + 430 + 242
∴ a = 245 kJ/mol
Q8. Heat of neutralization of a strong acid and a strong base is equal to ΔH of :
Solution:-
Heat of neutralisation is also defined as the heat of formation of H2O from H+ and OH− ions.
Heat of neutralisation is also defined as the heat of formation of H2O from H+ and OH− ions.
Q9. The enthalpy of vaporization of a liquid is 30 kJ/mol and entropy of vaporization is 75 J/mol. The boiling point of liquid at 1 atm is :
Solution:-
Tb = ΔH / ΔS = 400 K
Tb = ΔH / ΔS = 400 K
Q10. The apparatus generally used for measuring heat changes is:
Solution:-
Heat measurements are made in calorimeter usually made of copper.
Heat measurements are made in calorimeter usually made of copper.