Question

At what approximate temperature will an equimolar mixture of `N_2(g)` and `O_2(g)` be 3.0 % converted to `NO(g)`?

Oxides of nitrogen are produced in high-temperature combustion processes. The essential reaction is `N_2(g)+O_2(g)⇌2NO(g)`

At what approximate temperature will an equimolar mixture of `N_2(g)` and `O_2(g)` be 3.0 % converted to `NO(g)`?
[Hint: Use data from Appendix D in the textbook as necessary.]

Appendix D information... Please comment which part is needed because it's almost 7 pg.

Answer 1

Here's what I got.

Explanation:

The idea here is that you need to figure out the equilibrium constant that corresponds to a `3.0%` conversion to the product of an equimolar mixture of the two reactants.

Once you know the value of the equilibrium constant, you can use the Appendix to see at which temperature this takes place.

So, you know that you have

`"N"_ (2(g)) + "O"_ (2(g)) rightleftharpoons color(red)(2)"NO"_ ((g))`

By definition, the equilibrium constant for this equilibrium reaction takes the form

`K_c = (["NO"]^color(red)(2))/(["N"_2] * ["O"_2])`

Now, notice that every `1` mole of nitrogen gas that reacts consumes `1` mole of oxygen gas and produces `color(red)(2)` moles of nitrogen oxide.

You know that only `3.0%` of an equimolar mixture of nitrogen gas and oxygen gas must be converted to nitrogen oxide. If you take `x` to be the number of moles of nitrogen gas and of oxygen gas present in the mixture, you can say that, at equilibrium, the reaction vessel will contain

`overbrace(x color(white)(.)color(red)(cancel(color(black)("moles N"_2))) * (color(red)(2)color(white)(.)color(red)(cancel(color(black)("moles NO"))))/(1color(red)(cancel(color(black)("moles N"_2)))))^(color(purple)("= 100% conversion to produce 2"xcolor(white)(.)"moles NO")) * overbrace("3.0 moles NO"/(100color(red)(cancel(color(black)("moles NO")))))^(color(blue)("= 3.0% conversion")) = (0.06 * x)` `"moles NO"`

This means that the reaction consumed

`(0.06 * x) color(red)(cancel(color(black)("moles NO"))) * "1 mole N"_2/(color(red)(2)color(red)(cancel(color(black)("moles NO")))) = (0.03 * x)` `"moles N"_2`

and

`(0.06 * x) color(red)(cancel(color(black)("moles NO"))) * "1 mole O"_2/(color(red)(2)color(red)(cancel(color(black)("moles NO")))) = (0.03 * x)` `"moles O"_2`

You can thus say that, at equilibrium, you will have

`x color(white)(.)"moles N"_2 - (0.03 * x)color(white)(.)"moles N"_2 = (0.97 * x)` `"moles N"_2`

and

`x color(white)(.)"moles O"_2 - (0.03 * x)color(white)(.)"moles O"_2 = (0.97 * x)` `"moles O"_2`

All three gases share the same reaction vessel, which means that you can treat concentrations and number of moles interchangeably.

This means that the expression of the equilibrium constant becomes

`K_c = ( (0.06 * x)^color(red)(2))/( (0.97 * x) * (0.97 * x))`

This is equivalent to

`K_c = (0.06^color(red)(2) * color(red)(cancel(color(black)(x^color(red)(2)))))/(0.97^2 * color(red)(cancel(color(black)(x^2)))) = 0.003826`

Look up the temperature at which `K_c` is equal to the value we found and you'll get the answer.

Finally, the result makes sense because most of the equimolar mixture remains unreacted, so it makes sense to have `K_c < 1`, i.e. the equilibrium lies to the left.

Answer 2

see work below; edit if needed

Explanation:

This is the same problem, but some values might be different.

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`\DeltaG^0=\DeltaH^0-T\DeltaS^0`
`\DeltaH-T\DeltaS=\DeltaG^0=RT\lnk`

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`K=[NO]^2/([N_2][O_2]`
3% converted
`K=[.06]^2/([.97][.97])=3.83\cdot10^(-3)`

ICE table result `N_2=1.0-.03=.97`, `O_2=1.0-.03=.97`, `NO=0+0.06=0.06`

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`\DeltaH_f^0=9025 (kJ)/(mol)`
`\DeltaH_R=\Sigma_n\DeltaH_f^0(P)-\Sigma_n\DeltaH_f^0(R)`
`\DeltaH_f^0(O_2,NO_2)=0`

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`N_2(g)+O_2(g)\rightleftharpoons2NO(g)`
`\DeltaH_f^0-T\DeltaS^0=-RT\lnk`
`2(90.2(kJ)/(mol))-T(43.1\cdot10^(-3)(kJ)/K)=(-8.3145\cdot10^(-3)(kJ)/(mol K))(T)\ln(3.83\cdot10^(-3))`

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`\DeltaS^0=\SigmanS^0(P)=\SigmanS^0(R)`
`[2(219.9 J(mol K))]-[191.6J/(mol K)+205.1J/(mol K)]`
`=43.1J/K=43.1\cdot10^(-3)(kJ)/K`

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`0=\DeltaG=\DeltaG^0+RT\lnQ`
`\DeltaG^0=-RT\lnk`

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additional things

`T=(\DeltaH°)/(\DeltaS°-R\ln(K))`