# How does the vapor pressure in a vessel affects the total pressure after a combustion reaction?

The problem is as follows:

A sealed container of $$3\,L$$ in capactity is filled with hydrogen and oxygen at $$314\,torr$$ each one at a temperature of $$27^{\circ}C$$. A spark produces the reaction between oxygen and hydrogen. Find the total pressure after the reaction supposing that the temperature does not change. (You may use the vapor pressure of water at $$27^{\circ}C$$ is $$27\,torr$$)

The alternatives given are:

$$\begin{array}{ll} 1.&157\,torr\\ 2.&184\,torr\\ 3.&312\,torr\\ 4.&468\,torr\\ 5.&494\,torr\\ \end{array}$$

What I've attempted to do to solve this problem was to establish the initial conditions for the reaction:

$$O_2+2H_2\rightarrow 2H_{2}O$$

The number of moles for each gas can be found as follows: (for the sake of brevity I'm omitting units)

$$n_{O_{2}}=\frac{PV}{RT}=\frac{314\times 3}{62.4\times 300}= 0.0503$$

$$n_{H_{2}}=\frac{PV}{RT}=\frac{314\times 3}{62.4\times 300}= 0.0503$$

Both gases have the same number of moles.

From the reaction I could spot that the limiting reagent in this case is the hydrogen thus it will the one to be used for the calculation of the amount of water produced.

Since both hydrogen and water are in the same proportion the moles of water will be the same as what are the hydrogen.

$$n_{H_{2}O}=0.0503$$

While for the amount of oxygen which will remain will be half of the value since:

$$0.0503\,mol H_{2}\times\frac{1\,mol\,O_{2}}{2\,mol\,H_{2}}=0.0252\,mol\,O_{2}$$

Therefore this will be established as follows:

$$\begin{array}{llll} O_2&2H_2&\rightarrow&2H_{2}O\\ 1\,mol&2\,mol&&2\,mol\\ 0.0503&0.0503&&0\\ -0.0252&-0.0503&&+0.0503\\\hline -0.0251&-0.0503&&+0.0503\\ \end{array}$$

From this I went on to calculate the pressure for each:

For the water:

$$P=\frac{nRT}{V}=\frac{0.0503\times 62.4 \times 300}{3}= 313.82$$

For oxygen:

$$P=\frac{nRT}{V}=\frac{0.0251\times 62.4 \times 300}{3}= 156.62$$

But since it was given that the vapor pressure of water is $$27\,torr$$ made me confused on how should I use this given information. Does it mean that should I subtract from the oxygen?.

If I do that I obtain:

$$156.62-27=129.62$$

Then I assumed that this pressure can be added to the pressure of water:

$$313.82+129.62=443.44\,torr$$

But using this is way off from any of the alternatives in the given answers:

But If I add all the numbers:

$$313.82+156.62+27=497.44\,torr$$

Which is seems close to one of the alternatives but the number swapped. $$468\,torr$$

I'm not certain if this is due significant figures or an error in the problem. The source of my major confusion is exactly how should it be assesed the information regarding vapor pressure in a combustion reaction?. I'm assuming that after the combustion the water vapor formed will excert a pressure in the container so does this must be added to the pressure of the water formed resulting from the moles produced?.

Can somebody help me with this and clear out my doubts?.

First, kudos for actually showing work on a problem!

Second, I'd solve this by inspection, rather than using the Ideal Gas Law, where time is critical on a test.

As you correctly calculated, the limiting reactant is $$\ce{H2}$$. Therefore the pressure of $$\ce{H2}$$ is nil; all the $$\ce{H2}$$ is in the water.

Half of the $$\ce{O2}$$ is used, reacting with the $$\ce{H2}$$ to make $$\ce{H2O}$$, so the remaining partial pressure of $$\ce{O2}$$ is ~157 torr.

The volume taken up by the liquid water is ~1/2,000 that of the original gases, so the volume left in the flask is still ~3 l, though slightly less. Since the answer choices are not close together, and the decrease in volume is negligible, I'd ignore it.

The vapor pressure of water at 25° C is ~24 torr, and at 30° C it's ~32 torr, so at 27° C, it would be about halfway, by linear interpolation, say ~28 torr.

The sum of those partial pressures is close to an answer choice.

Trivia: This question seems to be citing pi with the initial value, 3[.]14 torr. If the temperature had been -8° C (ice at that temperature has a v.p. ~2 torr), the answer would have been 159 torr, so the first six digits of pi could have been used; i.e. 314 ->159.

• Interesting trivia. I totally overlooked that. However I'm still confused about the answer. Should it be $314+157+27=498$? (although it doesn't match any of those answers with 494 the only closest one). Btw the partial pressure of water at that temperature was given in the problem. I'm not sure if the water vapor pressure should be subtracted or added and the justification for that. Can you help me with this part?. – Chris Steinbeck Bell Jan 2 at 1:12
• @ChrisSteinbeckBell, from where did you get a partial pressure of 314 torr at the end? At the beginning, there was H2 and O2, but at the end, you state all the H2 was used up, as well as half the O2. – DrMoishe Pippik Jan 2 at 1:21
• I didn't noticed. So the water formed at that temperature became condensed the only gas remaining in the flask is oxygen and vapor pressure due water, hence $157+27=184$?. Is this alright now?. – Chris Steinbeck Bell Jan 2 at 1:25
• It looks good to me. Again, when taking multiple-choice tests, there are usually a few ridiculous answers, a close-but-not-quite-right choice, and the correct one. When pressed for time, see which of the two best answers make the most sense. – DrMoishe Pippik Jan 2 at 5:07