Given are the following data at 298 K. ΔfH∘SO2,g=−296.8kJmol−1 and ΔfH∘SO3,g=−395.7kJmol−1 The value of Li/.!° for the reaction SO2(g)+12O2(g)⟶SO3(g) is

A
$- 108.67 k J m o l^{- 1}$
B
$- 105.24 k J m o l^{- 1}$
C
$- 100.14 k J m o l^{- 1}$
D
$- 97.66 k J m o l^{- 1}$

Solution:

$(i) S (s) + O_{2} (g) ⟶ {SO}_{2} (g); Δ H^{\circ} = - 296.8 kJ {mol}^{- 1}$

$(ii) S (s) + \frac{3}{2} O_{2} (g) ⟶ {SO}_{3} (g); Δ H^{\circ} = - 395.7 kJ {mol}^{- 1}$

The required reaction ${SO}_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ {SO}_{3} (g)$ is obtained as Eq.(ii) - Eq. (i) Hence, $Δ_{r} H^{\circ} = [- 395.7 - (- 296.8)] kJ {mol}^{- 1} = - 98.9 kJ {mol}^{- 1}$ For the given reaction , $Δ V_{g} = - 1 / 2$ . Hence $\begin{aligned} Δ_{r} U^{\circ} = Δ_{r} H^{\circ} - (Δ v_{g}) R T = - 98.9 kJ {mol}^{- 1} - (- 1 / 2) (8.314 \times 10^{- 3} {kJK}^{- 1} {mol}^{- 1} (298 K) \\ = - (98.9 + 1.24) {kJmol}^{- 1} = - 97.66 kJ {mol}^{- 1} \end{aligned}$

Given are the following data at 298 K. $Δ_{f} H^{\circ} ({SO}_{2}, g) = - 296.8 kJ {mol}^{- 1} and Δ_{f} H^{\circ} ({SO}_{3}, g) = - 395.7 kJ {mol}^{- 1}$ The value of Li/.!° for the reaction ${SO}_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ {SO}_{3} (g) is$

Solution:

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