Short Answer TypeCalculate the enthalpy change of reaction:
at 298K, given that the enthalpies of formation at 298 K for 
Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4 (g) are -110, -393, 81 and 9.7 kJ mol-1. Find the value of 
for the reaction:
N2O4 (g) + 3CO(g) 
N2O(g) + 3CO2(g)
Long Answer TypeThe combustion of one mole of benzene takes place at 298K and 1 atm. After combustion, CO2(g) and H2O(l) are produced and 3267.0 kJ of heat is liberated. Calculate the standard enthalpy of formation of ∆fH° benzene, given that standard enthalpy of formation of CO2(g) and H2O(l) are –393.5 kJ mol–1 and –285.83 kJ mol–1respectively.
Short Answer TypeLong Answer TypeExplain why enthalpy of neutralization of a strong acid and strong base remains the same and the value changes if one of them is weak.
Enthalpy of neutralisation for a strong acid and a strong base is always constant: This is because in dilute solution all strong acids and strong bases are completely ionised. The neutralisation of a strong acid and strong base simply involves the combination of H+ions (from acid) and OH– ions (from base) to form unionised water molecules with the evolution of 57.1 kJ heat.
Since the same reaction takes place during neutralisation of all strong acids and strong bases, the value of enthalpy of neutralisation is constant.
The neutralisation of HCl and NaOH can be represented as:
Cancelling the common ions,
Enthalpy of neutralisation if either acid or base is weak : If one of the acid or bases is weak, then its ionisation is not complete in solution. Therefore, a part of the energy liberated is utilised for the ionisation of a weak acid (or base). Consequently, the value of enthalpy of neutralisation of the weak acid-strong base or strong acid-weak base is numerically less than 57.1 kJ. For example, neutralisation of acetic acid and sodium hydroxide can be represented as:
Thus, enthalpy of neutralisation of acetic acid and sodium hydroxide is –55.2 kJ.
Similarly, in the neutralisation of NH4OH and HCl, 5.6 kJ of heat is used up for the dissociation of weak base i.e. NH4OH. Hence enthalpy of neutralisation, in this case, is –57.1 + 5.6 = –51.5 kJ.
Explain the following:
(i) Enthalpy of fusion
(ii) Enthalpy of vaporization
(iii) Enthalpy of sublimation.
Short Answer Type
if the heats of formation of 
respectively.
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