Help with applying equilibrium constants?

[Swenis]

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Hello sirs and ladies 😉

I'm working on some gen chem 2 practice probs, and I've come across a problem that I can't figure out at all. My professor posted the solutions outside his classroom, but I can't pick them up because the school will not unlock the building grrrr. For some reason, these problems are giving me a fit even though I have all the formulas. I just can't manage to plug them in correctly. If anyone could help with this or explain how to do Keq problems, I'd be thankful 🙂

"At 2000 oC the equilibrium constant for the reaction 2NO(g) <-> N2(g) + O2(g) is Keq = 2.4E3. If the initial partial pressure of NO is 37.3 atm, what are the equilibrium partial pressures of NO, N2, and O2?"

Thanks 👍

 

[DrBowtie]

Try the MCAT forum.

 

[Swenis]

Sorry about posting in the wrong place. If a mod would move this to the MCAT forum, I'd appreciate it.

 

[TicAL]

You can just use an ICE table here like you would for any other similar equilibrium problem:

2 NO(g) --> N2(g) + O2(g)
Initial (37.3) (0) (0)
Change (-2x) (+x) (+x)
Equilibrium (37.3-2x) (+x) (+x)

Keq = products/reactants and equilibrium.

2.4E3 = (x)(x)/(37.3-2x)^2

Since Keq is relatively large, I don't think you can ignore the "-2x" portion of the denominator, so you'd have to solve the resulting quadratic formula to get a value for x. Then you just plug in the x value into the equilibrium portion of the table above to solve for all of the values you need.

*edit* the table looks kind of bad, but I put parenthesis around each of the values to help keep them seperate.

 

[Swenis]

You can just use an ICE table here like you would for any other similar equilibrium problem:

2 NO(g) --> N2(g) + O2(g)
Initial (37.3) (0) (0)
Change (-2x) (+x) (+x)
Equilibrium (37.3-2x) (+x) (+x)

Keq = products/reactants and equilibrium.

2.4E3 = (x)(x)/(37.3-2x)

Since Keq is relatively large, I don't think you can ignore the "-2x" portion of the denominator, so you'd have to solve the resulting quadratic formula to get a value for x. Then you just plug in the x value into the equilibrium portion of the table above to solve for all of the values you need.

*edit* the table looks kind of bad, but I put parenthesis around each of the values to help keep them seperate.

Thanks, I think I understand what you did, but it's been a long weekend and I don't remember how to know when to subtract x from each side or whether to add. Could you explain really quickly?

 

[TicAL]

Just think about what's happening. When you have something reacting to form products, you're going to lose some of the initial reactants while you'll create products.
Therefore, in equilibrium problems you always subtract from the reactants, and add to the products. You add and subtract based on the stochiometry of the problem (if you start with 2 moles you lose two moles, etc)

 

[obgyny]

okay, being the nerd I am, I just tried the problem. I just took Gen. Chem 2 last spring. I attached my explanation on a microsoft word thing. But I think I got the answers wrong because I tried plugging them back into the Keq equation, and they didn't equal 2.4E3.

TiCal-- did your anwers work out?

 

[Swenis]

Thanks guys, you're all awesome, and I figured it out thanks to you. I have found another question towards the end of the chapter I was wondering if you smart guys could help me with.

"NiO is to be reduced to nickel metal in an industrial process by use of the reaction

NiO(s) + CO(g) <-> Ni(s) + CO2(g)

At 1600 K the equilibrium constant for the reaction is Keq = 6.0E2. If a CO pressure of 150 torr is to be employed in the furnace and total pressure never exceeds 760 torr, will reduction occur?"

Thanks 🙂