It is known that anhydrous copper sulfate decomposes at high temperature according to the following two equations:

(1) $\mathrm{CuSO}_{4} \rightarrow \mathrm{CuO}(\mathrm{s})+\mathrm{SO}_{3}(\mathrm{g}) \quad K_{1}{ }^{\circ}$

(2) $2 \mathrm{SO}_{3}(\mathrm{g}) \rightarrow 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \quad K_{2}{ }^{\circ}$

1. Now, fill a $1.00 \mathrm{~L}$ closed container with argon gas at $273 \mathrm{~K}$ and $100 \mathrm{~mm} \mathrm{Hg}$ column pressure, and add a weighed amount of anhydrous copper sulfate solid. Heat to $1050 \mathrm{~K}$ and when the reaction reaches equilibrium, the mass of the solid decreases by $0.3869 \mathrm{~g}$, and the pressure in the closed container is $108.59 \mathrm{kPa}$. Calculate $K_{1}{ }^{\circ}$ and $K_{2}{ }^{\circ}$ at this temperature.

2. Assuming that the difference in standard molar reaction enthalpy between reaction (1) and reaction (2) is $\Delta_{\mathrm{r}} H_{\mathrm{m}}{ }^{\circ}(1)-\Delta_{\mathrm{r}} H_{\mathrm{m}}{ }^{\circ}(2)=27 \mathrm{~kJ} / \mathrm{mol}$, calculate the temperature at which the partial pressure of $\mathrm{SO}_{2}$ is 5 times the partial pressure of $\mathrm{SO}_{3}$ when equilibrium is reached.

3. At higher temperatures (>1400 K), the mass of the solid will further decrease. At the same certain temperature (>1400 K), due to the generation of new reactions, have the partial pressures of various gases increased or decreased compared to the original?

Answer the above questions and select the correct option.
已知无水硫酸铜在高温下按以下两式分解:

(1) $\mathrm{CuSO}_{4} \rightarrow \mathrm{CuO}(\mathrm{s})+\mathrm{SO}_{3}(\mathrm{g}) \quad K_{1}{ }^{\circ}$

(2) $2 \mathrm{SO}_{3}(\mathrm{g}) \rightarrow 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \quad K_{2}{ }^{\circ}$

1. 现将 $273 \mathrm{~K}$, $100 \mathrm{~mm} \mathrm{Hg}$ 柱压力的氩气充满 $1.00 \mathrm{~L}$ 的密闭容器, 放入称量好的无水硫酸铜固体。加热至 $1050 \mathrm{~K}$ 且反应达平衡时, 固体的质量减少了 $0.3869 \mathrm{~g}$, 密闭容器的压力为 $108.59 \mathrm{kPa}$ 。计算此温度下的 $K_{1}{ }^{\circ}$ 和 $K_{2}{ }^{\circ}$ 。

2. 假设反应(1)和反应(2)的标准摩尔反应焓之差 $\Delta_{\mathrm{r}} H_{\mathrm{m}}{ }^{\circ}(1)-\Delta_{\mathrm{r}} H_{\mathrm{m}}{ }^{\circ}(2)=27 \mathrm{~kJ} / \mathrm{mol}$, 计算当温度为多少时, 达平衡时 $\mathrm{SO}_{2}$ 的分压是 $\mathrm{SO}_{3}$ 分压的 $\mathrm{5}$ 倍。

3. 在更高的温度下(>1400 K),固体的质量会进一步减少。在相同的某温度下(>1400 K), 由于新反应的产生, 各种气体的分压与原来相比是增加了还是减少了。

回答以上问题，选出正确的选项。
All other options are incorrect
1. $K_{1}{ }^{\circ}=0.120$,
$K_{2}{ }^{\circ}=0.949 $

2. $T=6.3 \times 10^{2} \mathrm{~K} $

3. $p\left(\mathrm{O}_{2}\right)$ increases, $p\left(\mathrm{SO}_{2}\right)$ increases, $p\left(\mathrm{SO}_{3}\right)$ decreases
1. $K_{1}{ }^{\circ}=0.120$,
$K_{2}{ }^{\circ}=0.949 $

2. $T=6.3 \times 10^{2} \mathrm{~K} $

3. $p\left(\mathrm{O}_{2}\right)$ increases, $p\left(\mathrm{SO}_{2}\right)$ decreases, $p\left(\mathrm{SO}_{3}\right)$ decreases
1. $K_{1}{ }^{\circ}=0.120$,
$K_{2}{ }^{\circ}=0.949 $

2. $T=6.3 \times 10^{2} \mathrm{~K} $

3. $p\left(\mathrm{O}_{2}\right)$ increases, $p\left(\mathrm{SO}_{2}\right)$ decreases, $p\left(\mathrm{SO}_{3}\right)$ remains unchanged
1. $K_{1}{ }^{\circ}=0.120$,
$K_{2}{ }^{\circ}=0.949 $

2. $T=3.3 \times 10^{2} \mathrm{~K} $

3. $p\left(\mathrm{O}_{2}\right)$ increases, $p\left(\mathrm{SO}_{2}\right)$ decreases, $p\left(\mathrm{SO}_{3}\right)$ remains unchanged
1. $K_{1}{ }^{\circ}=0.120$,
$K_{2}{ }^{\circ}=0.949 $

2. $T=3.3 \times 10^{2} \mathrm{~K} $

3. $p\left(\mathrm{O}_{2}\right)$ increases, $p\left(\mathrm{SO}_{2}\right)$ increases, $p\left(\mathrm{SO}_{3}\right)$ increases
1. $K_{1}{ }^{\circ}=0.120$,
$K_{2}{ }^{\circ}=0.949 $

2. $T=3.3 \times 10^{2} \mathrm{~K} $

3. $p\left(\mathrm{O}_{2}\right)$ increases, $p\left(\mathrm{SO}_{2}\right)$ decreases, $p\left(\mathrm{SO}_{3}\right)$ increases
1. $K_{1}{ }^{\circ}=12.0$,
$K_{2}{ }^{\circ}=94.9 $

2. $T=6.3 \times 10^{2} \mathrm{~K} $

3. $p\left(\mathrm{O}_{2}\right)$ increases, $p\left(\mathrm{SO}_{2}\right)$ decreases, $p\left(\mathrm{SO}_{3}\right)$ remains unchanged
1. $K_{1}{ }^{\circ}=12.0$,
$K_{2}{ }^{\circ}=94.9 $

2. $T=3.3 \times 10^{2} \mathrm{~K} $

3. $p\left(\mathrm{O}_{2}\right)$ increases, $p\left(\mathrm{SO}_{2}\right)$ decreases, $p\left(\mathrm{SO}_{3}\right)$ decreases
1. $K_{1}{ }^{\circ}=12.0$,
$K_{2}{ }^{\circ}=94.9 $

2. $T=3.3 \times 10^{2} \mathrm{~K} $

3. $p\left(\mathrm{O}_{2}\right)$ increases, $p\left(\mathrm{SO}_{2}\right)$ decreases, $p\left(\mathrm{SO}_{3}\right)$ increases
其他选项均不正确
1. $K_{1}{ }^{\circ}=0.120$，
$K_{2}{ }^{\circ}=0.949 $

2. $T=6.3 \times 10^{2} \mathrm{~K} $

3. $p\left(\mathrm{O}_{2}\right)$ 变大，$p\left(\mathrm{SO}_{2}\right)$ 变大，$p\left(\mathrm{SO}_{3}\right)$ 变小
1. $K_{1}{ }^{\circ}=0.120$，
$K_{2}{ }^{\circ}=0.949 $

2. $T=6.3 \times 10^{2} \mathrm{~K} $

3. $p\left(\mathrm{O}_{2}\right)$ 变大，$p\left(\mathrm{SO}_{2}\right)$ 变小，$p\left(\mathrm{SO}_{3}\right)$ 变小
1. $K_{1}{ }^{\circ}=0.120$，
$K_{2}{ }^{\circ}=0.949 $

2. $T=6.3 \times 10^{2} \mathrm{~K} $

3. $p\left(\mathrm{O}_{2}\right)$ 变大，$p\left(\mathrm{SO}_{2}\right)$ 变小，$p\left(\mathrm{SO}_{3}\right)$ 不变
1. $K_{1}{ }^{\circ}=0.120$，
$K_{2}{ }^{\circ}=0.949 $

2. $T=3.3 \times 10^{2} \mathrm{~K} $

3. $p\left(\mathrm{O}_{2}\right)$ 变大，$p\left(\mathrm{SO}_{2}\right)$ 变小，$p\left(\mathrm{SO}_{3}\right)$ 不变
1. $K_{1}{ }^{\circ}=0.120$，
$K_{2}{ }^{\circ}=0.949 $

2. $T=3.3 \times 10^{2} \mathrm{~K} $

3. $p\left(\mathrm{O}_{2}\right)$ 变大，$p\left(\mathrm{SO}_{2}\right)$ 变大，$p\left(\mathrm{SO}_{3}\right)$ 变大
1. $K_{1}{ }^{\circ}=0.120$，
$K_{2}{ }^{\circ}=0.949 $

2. $T=3.3 \times 10^{2} \mathrm{~K} $

3. $p\left(\mathrm{O}_{2}\right)$ 变大，$p\left(\mathrm{SO}_{2}\right)$ 变小，$p\left(\mathrm{SO}_{3}\right)$ 变大
1. $K_{1}{ }^{\circ}=12.0$，
$K_{2}{ }^{\circ}=94.9 $

2. $T=6.3 \times 10^{2} \mathrm{~K} $

3. $p\left(\mathrm{O}_{2}\right)$ 变大，$p\left(\mathrm{SO}_{2}\right)$ 变小，$p\left(\mathrm{SO}_{3}\right)$ 不变
1. $K_{1}{ }^{\circ}=12.0$，
$K_{2}{ }^{\circ}=94.9 $

2. $T=3.3 \times 10^{2} \mathrm{~K} $

3. $p\left(\mathrm{O}_{2}\right)$ 变大，$p\left(\mathrm{SO}_{2}\right)$ 变小，$p\left(\mathrm{SO}_{3}\right)$ 变小
1. $K_{1}{ }^{\circ}=12.0$，
$K_{2}{ }^{\circ}=94.9 $

2. $T=3.3 \times 10^{2} \mathrm{~K} $

3. $p\left(\mathrm{O}_{2}\right)$ 变大，$p\left(\mathrm{SO}_{2}\right)$ 变小，$p\left(\mathrm{SO}_{3}\right)$ 变大
$n(\mathrm{Ar})=p_{0} V / R T_{0}=100 \times 1.00 \times 101.325 /(8.314 \times 760 \times 273)=5.87 \times 10^{-3}(\mathrm{~mol}) $ 

At $1050 \mathrm{~K}$, $p(\mathrm{Ar})=n R T / V=5.87 \times 10^{-3} \times 1050 \times 8.314 / 1.00=51.3(\mathrm{kPa}) $ 

<Checkpoint>[$p(\mathrm{Ar})=51.3\mathrm{kPa} $](1)</Checkpoint>

Therefore: $p\left(\mathrm{SO}_{2}\right)+p\left(\mathrm{O}_{2}\right)+p\left(\mathrm{SO}_{3}\right)=108.59 - 51.3=57.3(\mathrm{kPa})$

According to the conservation of mass,

$m\left(\mathrm{SO}_{2}\right)+m\left(\mathrm{O}_{2}\right)+m\left(\mathrm{SO}_{3}\right)=V\left[M\left(\mathrm{SO}_{2}\right) p\left(\mathrm{SO}_{2}\right)+M\left(\mathrm{O}_{2}\right) p\left(\mathrm{O}_{2}\right)+M\left(\mathrm{SO}_{3}\right) p\left(\mathrm{SO}_{3}\right)\right] / R T=0.3869 \mathrm{~g}$

$p\left(\mathrm{SO}_{2}\right)=2 p\left(\mathrm{O}_{2}\right) $

<Checkpoint>[$p\left(\mathrm{SO}_{2}\right)=2 p\left(\mathrm{O}_{2}\right) $](0.5)</Checkpoint>

Solving the three equations for three unknowns, we get:
$p\left(\mathrm{SO}_{2}\right)=30.2 \mathrm{kPa} $

$p\left(\mathrm{O}_{2}\right)=15.1 \mathrm{kPa} $

$p\left(\mathrm{SO}_{3}\right)=12.0 \mathrm{kPa}$

<Checkpoint>[$p\left(\mathrm{SO}_{2}\right)=30.2 \mathrm{kPa} $, $p\left(\mathrm{O}_{2}\right)=15.1 \mathrm{kPa} $, $p\left(\mathrm{SO}_{3}\right)=12.0 \mathrm{kPa}$](1.5)</Checkpoint>

Thus $K_{1}{ }^{\circ}=p\left(\mathrm{SO}_{3}\right) / p^{\circ}=0.120$

$K_{2}{ }^{\circ}=p^{2}\left(\mathrm{SO}_{2}\right) p\left(\mathrm{O}_{2}\right) / p^{\circ} p\left(\mathrm{SO}_{3}\right)=0.949 $

<Checkpoint>[$K_{1}{ }^{\circ}=0.120$, $K_{2}{ }^{\circ}=0.949 $](0.5)</Checkpoint>


The second question is based on the Van't Hoff equation:

$\ln \frac{K_{1}^{\mathrm{o}}(T)}{K_{1}^{\mathrm{o}}(1050 \mathrm{~K})}=\frac{\Delta_{r} H_{m}^{\mathrm{o}}(1)}{R}\left(\frac{1}{1050 \mathrm{~K}}-\frac{1}{T}\right) $

$\ln \frac{K_{2}^{\mathrm{o}}(T)}{K_{2}^{\mathrm{o}}(1050 \mathrm{~K})}=\frac{\Delta_{r} H_{m}^{\mathrm{o}}(2)}{R}\left(\frac{1}{1050 \mathrm{~K}}-\frac{1}{T}\right) $

Subtracting the two equations and rearranging slightly, we have:

$\ln \frac{K_{2}^{\mathrm{o}}(1050 K)}{K_{1}^{\mathrm{o}}(1050 K)}-\ln \frac{K_{2}^{\mathrm{o}}(T)}{K_{1}^{\mathrm{o}}(T)}=\frac{\Delta_{r} H_{m}^{\mathrm{o}}(1)-\Delta_{r} H_{m}^{\mathrm{o}}(2)}{R}\left(\frac{1}{1050 \mathrm{~K}}-\frac{1}{T}\right) $

According to the question, at this temperature:
$p\left(\mathrm{SO}_{2}\right)=5 p\left(\mathrm{SO}_{3}\right) $

In addition, under the initial conditions, $p\left(\mathrm{SO}_{2}\right)=2 p\left(\mathrm{O}_{2}\right) $. In reaction (2), the conversion rate of $\mathrm{SO}_{2}$ and the conversion rate of $\mathrm{O}_{2}$ are in a ratio of $2:1$, which is the same as the initial ratio. Therefore, the relationship $p\left(\mathrm{SO}_{2}\right)=2 p\left(\mathrm{O}_{2}\right) $ always holds.

Then:

$\frac{K_{2}^{\mathrm{o}}(T)}{K_{1}^{\mathrm{o}}(T)}=\frac{p_{\mathrm{SO}_{2}}{ }^{2} p_{\mathrm{O}_{2}}}{p_{\mathrm{SO}_{3}}{ }^{2}}=\frac{5^{2} \times \frac{5}{2}}{1}=62.5 $

Substituting all known data, we have

$\ln \frac{0.949}{0.120}-\ln 62.5=\frac{27 \times 10^{3}}{8.314}\left(\frac{1}{1050}-\frac{1}{T}\right) \quad$ 

Solving for $T$, we get $T=6.3 \times 10^{2} \mathrm{~K} $

<Checkpoint>[$T=6.3 \times 10^{2} \mathrm{~K} $](2)</Checkpoint>

At higher temperatures (>1400 K), the mass of the solid will further decrease, producing a new reaction. At this time, the only gas produced by the decomposition reaction can be oxygen. The chemical reaction equation for this step is:

$4 \mathrm{CuO} \rightarrow 2 \mathrm{Cu}_{2} \mathrm{O}+\mathrm{O}_{2} $ 

The new reaction occurring at this time is equivalent to adding extra oxygen to the original equilibrium. According to Le Chatelier's principle, the equilibrium of reaction (2) in the question shifts in the reverse direction, so $p\left(\mathrm{O}_{2}\right)$ increases and $p\left(\mathrm{SO}_{2}\right)$ decreases (these two substances are only related to the equilibrium of reaction (2)).

However, since the temperature remains constant, the equilibrium constant of the equilibrium remains unchanged, and $p\left(\mathrm{SO}_{3}\right)=K_{1}{ }^{\circ} p^{\circ}$, so $p\left(\mathrm{SO}_{3}\right)$ remains unchanged.

<Checkpoint>[$p\left(\mathrm{O}_{2}\right)$ increases](0.5)</Checkpoint>

<Checkpoint>[$p\left(\mathrm{SO}_{2}\right)$ decreases](0.5)</Checkpoint>

<Checkpoint>[$p\left(\mathrm{SO}_{3}\right)$ remains unchanged](0.5)</Checkpoint>
$n(\mathrm{Ar})=p_{0} V / R T_{0}=100 \times 1.00 \times 101.325 /(8.314 \times 760 \times 273)=5.87 \times 10^{-3}(\mathrm{~mol}) $ 

$1050 \mathrm{~K}$ 时, $p(\mathrm{Ar})=n R T / V=5.87 \times 10^{-3} \times 1050 \times 8.314 / 1.00=51.3(\mathrm{kPa}) $ 

<Checkpoint>[$p(\mathrm{Ar})=51.3\mathrm{kPa} $ ](1)</Checkpoint>

故：$p\left(\mathrm{SO}_{2}\right)+p\left(\mathrm{O}_{2}\right)+p\left(\mathrm{SO}_{3}\right)=108.59 - 51.3=57.3(\mathrm{kPa})$

根据物料守恒，

$m\left(\mathrm{SO}_{2}\right)+m\left(\mathrm{O}_{2}\right)+m\left(\mathrm{SO}_{3}\right)=V\left[M\left(\mathrm{SO}_{2}\right) p\left(\mathrm{SO}_{2}\right)+M\left(\mathrm{O}_{2}\right) p\left(\mathrm{O}_{2}\right)+M\left(\mathrm{SO}_{3}\right) p\left(\mathrm{SO}_{3}\right)\right] / R T=0.3869 \mathrm{~g}$

$p\left(\mathrm{SO}_{2}\right)=2 p\left(\mathrm{O}_{2}\right) $

<Checkpoint>[$p\left(\mathrm{SO}_{2}\right)=2 p\left(\mathrm{O}_{2}\right) $](0.5)</Checkpoint>

根据三个方程解三个未知数, 解得 ：
$p\left(\mathrm{SO}_{2}\right)=30.2 \mathrm{kPa} $

$p\left(\mathrm{O}_{2}\right)=15.1 \mathrm{kPa} $

$p\left(\mathrm{SO}_{3}\right)=12.0 \mathrm{kPa}$

<Checkpoint>[$p\left(\mathrm{SO}_{2}\right)=30.2 \mathrm{kPa} $, $p\left(\mathrm{O}_{2}\right)=15.1 \mathrm{kPa} $, $p\left(\mathrm{SO}_{3}\right)=12.0 \mathrm{kPa}$](1.5)</Checkpoint>

故 $K_{1}{ }^{\circ}=p\left(\mathrm{SO}_{3}\right) / p^{\circ}=0.120$

$K_{2}{ }^{\circ}=p^{2}\left(\mathrm{SO}_{2}\right) p\left(\mathrm{O}_{2}\right) / p^{\circ} p\left(\mathrm{SO}_{3}\right)=0.949 $

<Checkpoint>[$K_{1}{ }^{\circ}=0.120$, $K_{2}{ }^{\circ}=0.949 $](0.5)</Checkpoint>


第二问根据 Van't Hoff 方程：

$\ln \frac{K_{1}^{\mathrm{o}}(T)}{K_{1}^{\mathrm{o}}(1050 \mathrm{~K})}=\frac{\Delta_{r} H_{m}^{\mathrm{o}}(1)}{R}\left(\frac{1}{1050 \mathrm{~K}}-\frac{1}{T}\right) $

$\ln \frac{K_{2}^{\mathrm{o}}(T)}{K_{2}^{\mathrm{o}}(1050 \mathrm{~K})}=\frac{\Delta_{r} H_{m}^{\mathrm{o}}(2)}{R}\left(\frac{1}{1050 \mathrm{~K}}-\frac{1}{T}\right) $

两式相减, 并稍作变形有:

$\ln \frac{K_{2}^{\mathrm{o}}(1050 K)}{K_{1}^{\mathrm{o}}(1050 K)}-\ln \frac{K_{2}^{\mathrm{o}}(T)}{K_{1}^{\mathrm{o}}(T)}=\frac{\Delta_{r} H_{m}^{\mathrm{o}}(1)-\Delta_{r} H_{m}^{\mathrm{o}}(2)}{R}\left(\frac{1}{1050 \mathrm{~K}}-\frac{1}{T}\right) $

根据题意, 在此温度下:
$p\left(\mathrm{SO}_{2}\right)=5 p\left(\mathrm{SO}_{3}\right) $

另外，初始状态下$p\left(\mathrm{SO}_{2}\right)=2 p\left(\mathrm{O}_{2}\right) $，在反应(2)中，$\mathrm{SO}_{2}$转化率与$\mathrm{O}_{2}$转化率之比为$2:1$，与初始比例相同，因此$p\left(\mathrm{SO}_{2}\right)=2 p\left(\mathrm{O}_{2}\right) $的关系恒成立。

则：

$\frac{K_{2}^{\mathrm{o}}(T)}{K_{1}^{\mathrm{o}}(T)}=\frac{p_{\mathrm{SO}_{2}}{ }^{2} p_{\mathrm{O}_{2}}}{p_{\mathrm{SO}_{3}}{ }^{2}}=\frac{5^{2} \times \frac{5}{2}}{1}=62.5 $

代入所有已知数据,有

$\ln \frac{0.949}{0.120}-\ln 62.5=\frac{27 \times 10^{3}}{8.314}\left(\frac{1}{1050}-\frac{1}{T}\right) \quad$ 

解得 $T=6.3 \times 10^{2} \mathrm{~K} $

<Checkpoint>[$T=6.3 \times 10^{2} \mathrm{~K} $](2)</Checkpoint>

在更高的温度下(>1400 K),固体的质量会进一步减少, 产生新的反应，此时发生分解反应产生气体只能为氧气，此步反应的化学反应方程式为：

$4 \mathrm{CuO} \rightarrow 2 \mathrm{Cu}_{2} \mathrm{O}+\mathrm{O}_{2} $ 

此时发生新反应相当于在原平衡的基础上额外增加了氧气, 由勒沙特列原理, 题目中反应(2)的平衡逆向移动、故 $p\left(\mathrm{O}_{2}\right)$ 变大, $p\left(\mathrm{SO}_{2}\right)$ 变小(此两种物质只与反应(2)的平衡有关)。

但由于温度不变, 平衡的平衡常数不变, 而 $p\left(\mathrm{SO}_{3}\right)=K_{1}{ }^{\circ} p^{\circ}$, 故$p\left(\mathrm{SO}_{3}\right)$ 不变

<Checkpoint>[$p\left(\mathrm{O}_{2}\right)$ 变大](0.5)</Checkpoint>

<Checkpoint>[$p\left(\mathrm{SO}_{2}\right)$ 变小](0.5)</Checkpoint>

<Checkpoint>[$p\left(\mathrm{SO}_{3}\right)$ 不变](0.5)</Checkpoint>