September 29, 2020, 11:10:17 PM
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Topic: How can I find the total pressure in a container after a combustion of hydrogen?  (Read 408 times)

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Offline Blueberries116

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The problem is as follows:

A sealed container of [itex]3\,L[/itex] in capactity is filled with hydrogen and oxygen at [itex]314\,torr[/itex] each one at a temperature of [itex]27^{\circ}C.[/itex] 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 [itex]27^{\circ}C[/itex] is [itex]27\,torr[/itex])

The alternatives given are:

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

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

[itex]O_2+2H_2\rightarrow 2H_{2}O[/itex]

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

[itex]n_{O_{2}}=\frac{PV}{RT}=\frac{314\times 3}{62.4\times 300}= 0.0503[/itex]

[itex]n_{H_{2}}=\frac{PV}{RT}=\frac{314\times 3}{62.4\times 300}= 0.0503[/itex]

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.

[itex]n_{H_{2}O}=0.0503[/itex]

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

[itex]0.0503\,mol H_{2}\times\frac{1\,mol\,O_{2}}{2\,mol\,H_{2}}=0.0252\,mol\,O_{2}[/itex]

Therefore this will be established as follows:

[itex]\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}[/itex]

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

For the water:

[itex]P=\frac{nRT}{V}=\frac{0.0503\times 62.4 \times 300}{3}= 313.82[/itex]

For oxygen:

[itex]P=\frac{nRT}{V}=\frac{0.0251\times 62.4 \times 300}{3}= 156.62[/itex]

But since it was given that the vapor pressure of water is [itex]27\,torr[/itex] 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:

[itex]156.62-27=129.62[/itex]

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

[itex]313.82+129.62=443.44\,torr[/itex]

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

But If I add all the numbers:

[itex]313.82+156.62+27=497.44\,torr[/itex]

Which is seems close to one of the alternatives but the number swapped. [itex]468\,torr[/itex]

Can somebody help me with this?.
Have a nice day!

Offline Borek

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Do you think all produced water will stay in the gas phase at 27°C?
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Offline Blueberries116

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Do you think all produced water will stay in the gas phase at 27°C?

Then what will it happen?. Will it condense?. If so, should I subtract this from the pressure of water and from the unreacted oxygen?.

By doing this I'm getting:

[itex]313.82-27=286.82[/itex]

[itex]156.62-27=129.62[/itex]

[itex]286.82+129.62=416.44[/itex]

But again this answer doesn't get any close. What could it wrong?.
Have a nice day!

Offline Blueberries116

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Do you think all produced water will stay in the gas phase at 27°C?

It will condense. I'm assuming that the liquid compared to the total volume of the flask is very little so it can be negligible. But even after considering this, I'm confused on why my answer doesn't match?. Can you help me please ???
Have a nice day!

Offline Borek

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After condensation - what will be present in the gas phase? And what will be the pressure of each substance present?

Do you know Dalton's law?
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