2. (30) When $\mathrm{SO}_{3(\mathrm{~g})}$ is decomposed at the constant pressure $\mathbf{P}$ and $\mathrm{T}=1000 \mathrm{~K}$, the partial pressure of $\mathrm{O}_{2}$ in the equilibrium gas is $\mathbf{0 . 0 5} \mathrm{atm}$.
a) What is the pressure $\mathbf{P}$ ?
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 $\mathrm{pO}_{2}=0.05 \mathrm{~atm}$ ?
$$
\Delta \mathrm{G}^{\mathrm{o}}=-94600+89.37 \mathrm{~T} \text { for } \mathrm{SO}_{2}(\mathrm{~g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g})=\mathrm{SO}_{3}(\mathrm{~g})
$$

Question

2. (30) When $\mathrm{SO}_{3(\mathrm{~g})}$ is decomposed at the constant pressure $\mathbf{P}$ and $\mathrm{T}=1000 \mathrm{~K}$, the partial pressure of $\mathrm{O}_{2}$ in the equilibrium gas is $\mathbf{0 . 0 5} \mathrm{atm}$.
a) What is the pressure $\mathbf{P}$ ?
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 $\mathrm{pO}_{2}=0.05 \mathrm{~atm}$ ?
$$
\Delta \mathrm{G}^{\mathrm{o}}=-94600+89.37 \mathrm{~T} \text { for } \mathrm{SO}_{2}(\mathrm{~g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g})=\mathrm{SO}_{3}(\mathrm{~g})
$$

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Write the balanced decomposition reaction**: The decomposition of sulfur trioxide ($\mathrm{SO}_3$) into sulfur dioxide ($\mathrm{SO}_2$) and oxygen ($\mathrm{O}_2$) 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 ⋮ **Apply Dalton's Law of Partial Pressures**: According to Dalton's Law, the total pressure $P$ is the sum of the partial pressures of the individual gases. In this case, $P = p_{\mathrm{SO}_2} + p_{\mathrm{O}_2}$. Since the partial pressure of $\mathrm{O}_2$ is given as $0.05 \mathrm{~atm}$ and it is produced from $\mathrm{SO}_3$ in a 1:0.5 ratio, the partial pressure of $\mathrm{SO}_2$ will be twice that of $\mathrm{O}_2$, which is $0.10 \mathrm{~atm}$. Therefore, $P = 0.10 \mathrm{~atm} + 0.05 \mathrm{~atm} = 0.15 \mathrm{~atm}$. ‖ ‖ 3 ⋮ **Use the Gibbs free energy equation**: The temperature change at constant pressure can be found by using the Gibbs free energy equation $\Delta G = \Delta G^\circ + RT \ln Q$, where $Q$ is the reaction quotient. At equilibrium, $\Delta G = 0$, and $Q$ becomes the equilibrium constant $K_p$. We can rearrange the equation to solve for $T$ when the pressure is changed to $1.5 \mathrm{~atm}$ and $p_{\mathrm{O}_2}$ remains $0.05 \mathrm{~atm}$. ‖ ∻Answer∻ ⚹ a) The pressure $P$ is $0.15 \mathrm{~atm}$. b) To find the new temperature $T$, we need more information about the reaction, such as the equilibrium constant at the initial conditions or the change in the number of moles of gas during the reaction, which affects the reaction quotient $Q$. Without this information, we cannot calculate the exact temperature. ⚹ ∻Key Concept∻ ⚹ Dalton's Law of Partial Pressures ⚹ ∻Explanation∻ ⚹ Dalton's Law states that the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of individual gases. ⚹ --- For the second part of the question, we would need additional information to provide a complete answer. However, since the question does not provide enough data to calculate the new temperature after the pressure change, we can only state that the calculation would involve the reaction quotient and the Gibbs free energy equation.◊What is the relationship between the equilibrium constant and the change in Gibbs free energy for the reaction $SO_2(g) + \frac{1}{2}O_2(g) \rightleftharpoons SO_3(g)$?⍭ Generate me a similar question◊

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