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Q.a. Define entropy. A swimmer coming out from a pool is covered with a film of water weighing about 18g. How much heat must be supplied to evaporate this water at 298K. Calculate the internal energy of vaporization at 298K. OR b. Calculate the lattice enthalpy of Na+Cl- by Born Haber cycle.

Nagaland NbseNagaland Board of School Education (Class XI) 2021Subjective· 5mImportance★★★★★
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(a) Entropy measures disorder; using the molar enthalpy of vaporization of water, 1 mole (18 g) needs about 44.0 kJ to evaporate, and subtracting the ΔngRT\Delta n_g RT work term gives an internal energy of vaporization of about 41.5 kJ. (b) A Born–Haber cycle sums sublimation, ionization, bond dissociation, electron-gain and formation enthalpies to back out the lattice enthalpy of NaCl, about −787 kJ/mol.

(a) Entropy and evaporation of water

Entropy (S) is a thermodynamic state function that measures the degree of disorder or randomness of a system; a more disordered state has higher entropy. For a reversible process at constant temperature, ΔS=qrev/T\Delta S = q_{rev}/T.

Mass of water =18 g= 18\ \text{g}, molar mass =18 g mol−1= 18\ \text{g mol}^{-1}, so n=1 moln = 1\ \text{mol}.

Using the standard enthalpy of vaporization of water near room temperature, ΔvapH∘≈44.0 kJ mol−1\Delta_{vap}H^\circ \approx 44.0\ \text{kJ mol}^{-1} at 298 K:

q=n×ΔvapH∘=1×44.0=44.0 kJq = n \times \Delta_{vap}H^\circ = 1 \times 44.0 = 44.0\ \text{kJ}

Internal energy of vaporization: since ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RT (liquid → gas, Δng=+1\Delta n_g = +1 mol),

ΔU=ΔH−ΔngRT=44.0 kJ−(1)(8.314 J mol−1K−1)(298 K)=44.0−2.48=41.5 kJ\Delta U = \Delta H - \Delta n_g RT = 44.0\ \text{kJ} - (1)(8.314\ \text{J mol}^{-1}\text{K}^{-1})(298\ \text{K}) = 44.0 - 2.48 = 41.5\ \text{kJ}

(b) OR — Lattice enthalpy of NaCl by Born–Haber cycle

By Hess's law, the enthalpy of formation of NaCl(s) from its elements can be broken into steps:

ΔfH∘(NaCl)=ΔsubH∘(Na)+IE1(Na)+12ΔbondH∘(Cl2)+ΔegH∘(Cl)+ΔlatticeH∘(NaCl)\Delta_fH^\circ(\text{NaCl}) = \Delta_{sub}H^\circ(\text{Na}) + IE_1(\text{Na}) + \tfrac{1}{2}\Delta_{bond}H^\circ(\text{Cl}_2) + \Delta_{eg}H^\circ(\text{Cl}) + \Delta_{lattice}H^\circ(\text{NaCl}) …

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