Chemistry: 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

Lithium: 0.98beryllium: 1.57boron:2.04carbon:2.55nitrogen:3.04lithium is the answer. it means that the atoms in the left side of the periodic table have the lowest electronegativity value and the right side atoms have the highest hope this helps...

The value of δg° at 25 °c for the decomposition

Ответ:

mattotto1758

10.01.2022

Chemistry

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

Option: A

Explanation:

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

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

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

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

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