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vlada-n [284]
4 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]4 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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Applying Gay Lussac's law:

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Answer:

0.147 billion years = 147.35 million years.

Explanation:

  • It is known that the decay of a radioactive isotope isotope obeys first order kinetics.
  • Half-life time is the time needed for the reactants to be in its half concentration.
  • If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).
  • Also, it is clear that in first order decay the half-life time is independent of the initial concentration.
  • The half-life of Potassium-40 is 1.25 billion years.

  • For, first order reactions:

<em>k = ln(2)/(t1/2) = 0.693/(t1/2).</em>

Where, k is the rate constant of the reaction.

t1/2 is the half-life of the reaction.

∴ k =0.693/(t1/2) = 0.693/(1.25 billion years) = 0.8 billion year⁻¹.

  • Also, we have the integral law of first order reaction:

<em>kt = ln([A₀]/[A]),</em>

<em></em>

where, k is the rate constant of the reaction (k = 0.8 billion year⁻¹).

t is the time of the reaction (t = ??? year).

[A₀] is the initial concentration of (Potassium-40) ([A₀] = 100%).

[A] is the remaining concentration of (Potassium-40) ([A] = 88.88%).

  • At the time needed to be determined:

<em>8 times as many potassium-40 atoms as argon-40 atoms. Assume the argon-40 only comes from radioactive decay.</em>

  • If we start with 100% Potassium-40:

∴ The remaining concentration of Potassium-40 ([A] = 88.88%).

and that of argon-40 produced from potassium-40 decayed = 11.11%.

  • That the ratio of (remaining Potassium-40) to (argon-40 produced from potassium-40 decayed) is (8: 1).

∴ t = (1/k) ln([A₀]/[A]) = (1/0.8 billion year⁻¹) ln(100%/88.88%) = 0.147 billion years = 147.35 million years.

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<h3>Pacing Study Sessions</h3>

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What happens is that Lauren is using mass repetition processing, which can be compared to a longitudinal wave in physics, with spaces in between, concentrating the initial review close to the proof to ensure retention and avoid forgetting.

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