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

(i) S(s)+O2 (g) ⟶SO2(g); ΔH∘=−296.8kJmol−1 (ii) S(s)+32O2(g)⟶SO3(g); ΔH∘=−395.7kJmol−1The required reaction SO2(g)+12O2(g)⟶SO3(g) is obtained as Eq.(ii) - Eq. (i) Hence, ΔrH∘=[−395.7−(−296.8)]kJmol−1=−98.9kJmol−1 For the given reaction ,ΔVg=−1/2. Hence ΔrU∘=ΔrH∘−ΔvgRT=−98.9kJmol−1−(−1/2)8.314×10−3kJK−1mol−1(298K)=−(98.9+1.24)kJmol−1=−97.66kJmol−1

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

(i) S(s)+O2 (g) ⟶SO2(g); ΔH∘=−296.8kJmol−1 (ii) S(s)+32O2(g)⟶SO3(g); ΔH∘=−395.7kJmol−1The required reaction SO2(g)+12O2(g)⟶SO3(g) is obtained as Eq.(ii) - Eq. (i) Hence, ΔrH∘=[−395.7−(−296.8)]kJmol−1=−98.9kJmol−1 For the given reaction ,ΔVg=−1/2. Hence ΔrU∘=ΔrH∘−ΔvgRT=−98.9kJmol−1−(−1/2)8.314×10−3kJK−1mol−1(298K)=−(98.9+1.24)kJmol−1=−97.66kJmol−1

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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 ${SO}_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ {SO}_{3} (g) is$

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$- 97.66 k J m o l^{- 1}$

$- 100.14 k J m o l^{- 1}$

$- 105.24 k J m o l^{- 1}$

$- 108.67 k J m o l^{- 1}$

answer is D.

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Detailed 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 $\begin{aligned} Δ_{r} U^{\circ} = Δ_{r} H^{\circ} - (Δ v_{g}) R T = - 98.9 kJ {mol}^{- 1} - (- 1 / 2) (8.314 \times 10^{- 3} {kJK}^{- 1} {mol}^{- 1} (298 K) \\ = - (98.9 + 1.24) {kJmol}^{- 1} = - 97.66 kJ {mol}^{- 1} \end{aligned}$