When SO3 $(\mathrm{g})$ is decomposed at the constant pressure $\mathrm{P}$ and $\mathrm{T}=1000 \mathrm{~K}$, the partial pressure of $\mathrm{O} 2$ in the equilibrium gas is $0.05 \mathrm{~atm}$. a) What is the pressure $\mathrm{P}$ ? b) If the pressure of this equilibrated gas is increased to $1.5 \mathrm{~atm}$, to what value must be the temperature to decreased to produce a gas mixture in which $\mathrm{pO} 2=0.05 \mathrm{~atm} ? \Delta \mathrm{Go}=$ $-94600+89.37$ T for SO2 $(\mathrm{g})+1 / 2 \mathrm{O} 2(\mathrm{~g})=\mathrm{SO} 3(\mathrm{~g})$

Question

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Identify the balanced chemical reaction**: The decomposition of $\mathrm{SO}_3$ into $\mathrm{SO}_2$ and $\mathrm{O}_2$ can be represented by the balanced chemical equation: $\mathrm{SO}_3(g) \rightarrow \mathrm{SO}_2(g) + \frac{1}{2} \mathrm{O}_2(g)$. ‖ ‖ 2 ⋮ **Use the stoichiometry of the reaction to find total pressure**: Given that the partial pressure of $\mathrm{O}_2$ is $0.05 \mathrm{~atm}$, and according to the stoichiometry of the reaction, for every 1 mole of $\mathrm{SO}_3$ decomposed, 0.5 moles of $\mathrm{O}_2$ are produced, the partial pressure of $\mathrm{SO}_2$ will be twice that of $\mathrm{O}_2$, which is $0.10 \mathrm{~atm}$. The total pressure $\mathrm{P}$ is the sum of the partial pressures of $\mathrm{SO}_2$ and $\mathrm{O}_2$, so $\mathrm{P} = 0.10 \mathrm{~atm} + 0.05 \mathrm{~atm} = 0.15 \mathrm{~atm}$. ‖ ‖ 3 ⋮ **Apply Le Chatelier's principle for temperature change**: When the pressure of the equilibrated gas is increased, the equilibrium will shift to minimize this change. Since the reaction involves a decrease in the number of moles of gas, increasing the pressure will shift the equilibrium to the right, favoring the formation of $\mathrm{SO}_3$. To maintain the partial pressure of $\mathrm{O}_2$ at $0.05 \mathrm{~atm}$, the temperature must be decreased to shift the equilibrium back to the left. ‖ ‖ 4 ⋮ **Use the Gibbs free energy equation to find the new temperature**: The Gibbs free energy change ($\Delta \mathrm{G}$) for the reaction at equilibrium is zero. We can use the given $\Delta \mathrm{G}^\circ$ equation, $\Delta \mathrm{G}^\circ = -94600 + 89.37 \mathrm{T}$, to find the temperature at which the partial pressure of $\mathrm{O}_2$ remains $0.05 \mathrm{~atm}$ at the new pressure. Since $\Delta \mathrm{G} = \Delta \mathrm{G}^\circ + \mathrm{RT} \ln Q$, and at equilibrium $\Delta \mathrm{G} = 0$ and $Q = K$, we can set $\Delta \mathrm{G}^\circ = -\mathrm{RT} \ln K$. We need to solve for $\mathrm{T}$ when $\Delta \mathrm{G}^\circ = -94600 + 89.37 \mathrm{T}$ and $K$ corresponds to the equilibrium constant at the new pressure of $1.5 \mathrm{~atm}$. ‖ ∻1 Answer∻ ⚹ The pressure $\mathrm{P}$ is $0.15 \mathrm{~atm}$. ⚹ ∻2 Answer∻ ⚹ The temperature must be decreased to maintain $\mathrm{pO}_2 = 0.05 \mathrm{~atm}$, but the exact value requires additional information about the equilibrium constant at the new pressure. ⚹ ∻Key Concept∻ ⚹ Le Chatelier's principle and the relationship between Gibbs free energy and equilibrium ⚹ ∻Explanation∻ ⚹ Le Chatelier's principle predicts the direction of the shift in equilibrium when external conditions such as pressure and temperature change. The Gibbs free energy equation relates the standard free energy change to the temperature and the equilibrium constant. ⚹◊What is the equilibrium constant expression for the decomposition of SO3 at the given temperature and pressure?⍭ Generate me a similar question◊

Sia