Question 1. When liquid benzene is oxidised at constant pressure at 300 K, the change in enthalpy is -3728 kJ. What is the change in internal energy at the same temperature?
Question 3. Calculate the enthalpy change for the reaction: H2(g) + Cl2(g) ————-> 2HCl(g). Given that bond energies ofH-H, Cl- Cl and H-Cl bonds are 433, 244 and 431 kj mol-1 respectively.
Answer: The chemical equation for the reaction is:
H2(g) + Cl2(g) ———-> 2HCl(g)
The enthalpy of reaction is:
∆rH =∑B.E. of reactants – ∑B.E. of products= [B.E. of H-H bond + B.E. of Cl-Cl bond]
– [2 x B.E. of H—Cl bond]
= (433 + 244) – (2 x 431) = 433 + 244 – 862 = -185 kj
Question 4. The bond enthalpy of H2(g) is 436 kj mol-1and that of N2 (g) is 941.3 kj mol-1. Calculate the average bond enthalpy of an N-H bond in ammonia. Given: ∆H– (NH3) = -46 kj mol-1
Answer:
Question 12. Give reason for the following:
(a)Neither q nor w is a state function but q + w is a state function.
(b)A real crystal has more entropy than an ideal crystal.
Answer: (a) q + w = ∆u
As ∆u is a state function hence, q + w is a state function.
(b) A real crystal has some disorder due to the presence of defects in its structural arrangement whereas ideal crystal does not have any disorder. Hence, a real crystal has more entropy than an ideal crystal.
