InterviewSolution
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InterviewSolution
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(a) What is standard enthalpy of formation?(b) GivenN2 (g) + 3H2 (g) → 2NH3 (g); ∆rH0 = -98.4 kJ/mol.What is the standard enthalpy of formation of ammonia gas?(c) Differentiate between intensive property and extensive property.(d) Enthalpies of formation of CO(g), CO2 (g),N2O(g), and N2O4(g) are -110, 393, 81 and 9.7 kJ/mol respectively. Find the value of ∆rH for the reaction N2O4 (g) + 3CO(g) → N2O(g) + 3CO2(g) |
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Answer» (a) Standard enthalpy of formation (∆fH° ) is enthalpy change accompanying the formation of 1 mole of the substance from its constituent elements in their standard state. ∆fH° can be > 0 or < 0. (b) Given: N2 (g) + 3H2(g) → 2NH3 (g); = ∆fH° = −92.4 kJ mol−1 This is heat evolved for 2 moles of NH3 (g) \(\therefore\) Heat evolved for 1 mole of NH3 (g) = \(\frac{-92.4}{2}\) = -46.2 kJ Hence ∆fH° of NH3 gas = - 46.2 kJ mol−1 The properties that depend on the quantity of matter contained in the system are called extensive properties. For example, mass, volume etc. On the other hand, the properties which depend on the nature of the substance and not on the amount of substance are called intensive properties. For example, viscosity etc. (d) N2O4 (g) + 3CO(g) → N2O (g) + 3CO2 (g) ΔrH = ? ∆rH =\(\sum\)∆fH(product) - \(\sum\)∆fH(reactants) = [∆fH(N2O) + 3 × ∆fHCO2 ] − [∆fH(N2O4 ) + 3 × ∆fCO] = [81 + 3 × (−393)] − [9.7 + 3(−110)] = −777.7 kJ/mol |
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