2SO2(g) + O2(g) - 2SO3(g)

The above reaction is carried out in a vessel starting with partial pres

Q: Why are we treating partial pressures like moles?

According to the Ideal Gas Law ($P = \frac{nRT}{V}$), at constant temperature and volume, partial pressure is directly proportional to the number of moles. Therefore, we can use pressures directly in our stoichiometric calculations!

Q: Why did we divide 250 by 2 for oxygen's change?

Look at the balanced equation: 2 moles of $\text{SO}_2$ react with 1 mole of $\text{O}_2$. This means the amount of $\text{O}_2$ consumed is exactly half the amount of $\text{SO}_2$ consumed. Since 250 mbar of $\text{SO}_2$ reacted, $\frac{250}{2} = 125$ mbar of $\text{O}_2$ must react.

Q: Why isn't this an equilibrium problem if it's in the Chemical Equilibrium chapter?

Great observation! The key phrase in the question is 'When the reaction is complete'. This implies a 100% yield where the reaction goes to completion, making it a classic limiting reagent stoichiometry problem rather than an equilibrium one.

Q: What if $\text{O}_2$ was the limiting reagent?

If $\text{O}_2$ were limiting, all 750 mbar of it would be consumed. Then, we would need $750 \times 2 = 1500$ mbar of $\text{SO}_2$ to react with it. Since we only have 250 mbar of $\text{SO}_2$, it's impossible for $\text{O}_2$ to be the limiting reagent here.

Q: Does the total pressure always decrease in this reaction?

Yes! Notice that 3 moles of reactant gases (2 $\text{SO}_2$ + 1 $\text{O}_2$) are converting into 2 moles of product gas (2 $\text{SO}_3$). Since the total number of gas molecules decreases, the total pressure in a rigid vessel will also decrease as the reaction proceeds.

Question

$2	ext{SO}_2(g) + 	ext{O}_2(g) \longrightarrow 2	ext{SO}_3(g)$

The above reaction is carried out in a vessel starting with partial pressure $p_{	ext{SO}_2} = 250 	ext{ m bar}$, $p_{	ext{O}_2} = 750 	ext{ m bar}$ and $p_{	ext{SO}_3} = 0$. When the reaction is complete, the total pressure in the reaction vessel is ............ m bar. (Round off of the nearest integer).

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine you are a chemical engineer tasked with maximizing the production of sulfur trioxide ($ 	ext{SO}_3 $). You pump sulfur dioxide ($ 	ext{SO}_2 $) and oxygen ($ 	ext{O}_2 $) into a rigid reaction vessel.

The problem gives us the initial partial pressures of the gases: $250 	ext{ m bar}$ for $ 	ext{SO}_2 $ and $750 	ext{ m bar}$ for $ 	ext{O}_2 $. Initially, there is no $ 	ext{SO}_3 $ present.

The most crucial piece of information here is the phrase **"When the reaction is complete"**. This tells us that we are not dealing with a delicate equilibrium state. Instead, the reaction will proceed forward relentlessly until one of the reactants is completely exhausted.

### The Master Equation and Limiting Reagent
To find out which gas runs out first, we need to identify the **limiting reagent**. We do this by comparing the ratio of the initial amount (or in this case, partial pressure) to the stoichiometric coefficient for each reactant.

For $ 	ext{SO}_2 $, the ratio is $ \frac{250}{2} = 125 $. 
For $ 	ext{O}_2 $, the ratio is $ \frac{750}{1} = 750 $.

Since $125$ is significantly less than $750$, $ 	ext{SO}_2 $ is our limiting reagent. It will dictate the pace and the ultimate yield of the entire reaction. Every single molecule of $ 	ext{SO}_2 $ will be consumed!

### Tracking the Changes
Now, let's map out the changes in pressure. Since $ 	ext{SO}_2 $ is the limiting reagent, its pressure will decrease by exactly $250 	ext{ m bar}$.

According to the balanced chemical equation, $2$ moles of $ 	ext{SO}_2 $ react with $1$ mole of $ 	ext{O}_2 $. Therefore, the amount of $ 	ext{O}_2 $ consumed will be exactly half of the $ 	ext{SO}_2 $ consumed. 
The change in $ 	ext{O}_2 $ pressure is $ -\frac{250}{2} = -125 	ext{ m bar} $.

What about our product? The stoichiometry shows that $2$ moles of $ 	ext{SO}_2 $ produce $2$ moles of $ 	ext{SO}_3 $. This $1:1$ ratio means the pressure of $ 	ext{SO}_3 $ will increase by the exact same amount that $ 	ext{SO}_2 $ decreased. 
The change in $ 	ext{SO}_3 $ pressure is $ +250 	ext{ m bar} $.

### Final Calculation
Let's tally up the final state of our reaction vessel. 
The $ 	ext{SO}_2 $ is completely gone, leaving $0 	ext{ m bar}$. 
The $ 	ext{O}_2 $ started at $750 	ext{ m bar}$ and lost $125 	ext{ m bar}$, leaving us with a final pressure of $625 	ext{ m bar}$. 
The $ 	ext{SO}_3 $ started at $0$ and gained $250 	ext{ m bar}$, ending at $250 	ext{ m bar}$.

The question asks for the **total pressure** in the vessel. According to Dalton's Law of Partial Pressures, we simply sum the individual pressures of all gases present at the end.

$$ P_{	ext{total}} = P_{	ext{SO}_2} + P_{	ext{O}_2} + P_{	ext{SO}_3} $$
$$ P_{	ext{total}} = 0 + 625 + 250 = 875 	ext{ m bar} $$

And there we have it! The final total pressure in the vessel is $875 	ext{ m bar}$.

Enter Numerical Value:

Visualized Solution

$2 SO_{2} + O_{2} ⟶ 2 SO_{3}$

$Initial Partial Pressures$

$Identifying Limiting Reagent$

$Limiting Reagent Conclusion$

$Change in Reactants$

$Change in Product$

$Final Partial Pressures$

$Dalton’s Law of Partial Pressures$

$Total Final Pressure$

$Equilibrium vs Completion$

The Sigma Insight: Stoichiometric and Volumetric Calculations

Analyzing the Setup Imagine you are a chemical engineer tasked with maximizing the production of sulfur trioxide ( $SO_{3}$ )

The Master Equation and Limiting Reagent To find out which gas runs out first, we need to identify the limiting reagent

Tracking the Changes Now, let's map out the changes in pressure

Final Calculation Let's tally up the final state of our reaction vessel

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2SO2​+O2​⟶2SO3​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial Partial Pressures

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Identifying Limiting Reagent

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Limiting Reagent Conclusion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Change in Reactants

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Change in Product

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Partial Pressures

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Dalton’s Law of Partial Pressures

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Final Pressure

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equilibrium vs Completion