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vlada-n [284]
3 years ago
11

The value of ΔG° at 25 °C for the decomposition of gaseous sulfur trioxide to solid elemental sulfur and gaseous oxygen, 2SO3 (g

) → 2S (s, rhombic) + 3O2 (g) is ________ kJ/mol. The value of G° at 25 °C for the decomposition of gaseous sulfur trioxide to solid elemental sulfur and gaseous oxygen, 2SO3 (g) 2S (s, rhombic) + 3O2 (g) is ________ kJ/mol. +740.8 -370.4 +185.2 +370.4 -740.8
Chemistry
1 answer:
Zarrin [17]3 years ago
4 0

Answer:

\begin{array}{l}{\text { The value of } \Delta \mathrm{G}^{\circ} \text { at } 25^{\circ} \mathrm{C} \text { for the decomposition of gaseous sulfur trioxide to solid }} \\ {\text { elemental sulfur and gaseous oxygen is }+740.8 \mathrm{kJ} / \mathrm{mol}}\end{array}

Option: A

<u>Explanation</u>:

\begin{array}{l}{\text { The value of } \Delta \mathrm{G}^{\circ} \text { at } 25^{\circ} \mathrm{C} \text { in the following reaction can be calculated as follows: }} \\ {2 \mathrm{SO}_{3}(\mathrm{g}) \rightarrow 2 \mathrm{S}(\mathrm{s}, \text { rhombic })+3 \mathrm{O}_{2}(\mathrm{g})}\end{array}

\begin{array}{l}{\Delta \mathrm{G}^{\circ} \text { is Standard Gibbs free energy change which can be calculated from the standard free }} \\ {\text { energies of formation of the products and the reactants from the following equation: }}\end{array}

\begin{array}{l}{\Delta \mathrm{G}^{\circ}=\Sigma \mathrm{G}_{\mathrm{f}(\text { products })}^{\circ}-\Sigma \mathrm{G}_{\text {creatants }}^{\circ}} \\ {\Delta \mathrm{G}^{\circ}=[\mathrm{Sum} \text { of standard free energies of formation of products }]-[\mathrm{Sum}\text { of standard } } \\ {\text { free energies bf formation of reactants] }}\end{array}

\begin{array}{l}{\text { Now here standard values of } \Delta G^{\circ} f(k J / m o l) \text { for } S=0, O_{2}=0 \& S O_{3}=-370.4} \\ {\text { Hence these values can be substituted in above equation: }} \\ {\Delta G^{\circ}=\left[2 G_{f}^{\circ}(0)+3 G_{f}(0)\right]-[2(-370.4)]} \\ {\Delta G^{\circ}=[-0+0]-[-740.8]} \\ {\Delta G^{\circ}=+740.8 \mathrm{kJ} / \mathrm{mol}}\end{array}

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7. Disulfur dichloride can be made by reacting chlorine gas with molten sulfur. What is the yield of S2Cl2 expected in a laborat
Pie

Answer:

11.4g of S₂Cl₂ is the expected yield

9.69g of S₂Cl₂ are produced with a 85% yield

Explanation:

The reaction of sulfur S₈ with Cl₂ to produce S₂Cl₂ is:

S₈ + 4Cl₂ → 4S₂Cl₂

<em>Where 1 mole of sulfur reacts with four moles of chlorine to produce four moles of disulfur dichloride.</em>

To find the limiting reactant you need to convert mass of each reactant to moles using its molar mass, thus:

S₈ (Molar mass: 256.52g/mol): 10.0g ₓ (1mol / 256.52g) = 0.0390 moles S₈

Cl₂ (Molar mass: 70.9g/mol): 6.00g ₓ (1mol / 70.9g) = 0.0846 moles Cl₂

For a complete reaction of 0.0390 moles of sulfur, there are necessaries:

0.0390 mol S₈ ₓ (4 mol Cl₂ / 1 mol S₈) = <em>0.156 moles Cl₂. </em>As you have just 0.0846 moles of chlorine, Cl₂ is the limiting reactant.

As 4 moles of Cl₂ produce 4 moles of S₂Cl₂.<em> 0.0846 moles of Cl₂ produce, in theory, 0.0846 moles of S₂Cl₂ (Molar mass: 135.04g/mol). </em>In mass:

0.0846 moles S₂Cl₂ ₓ (135.04g/mol) =

<h3>11.4g of S₂Cl₂ is the expected yield</h3>

If you produce just the 85.0% of yield, mass of S₂Cl₂ is:

11.4g ₓ 85% =

<h3>9.69g of S₂Cl₂</h3>
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3 years ago
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