Chemistry question on DAT Achiever

[Tracy47]

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Could anyone tell me why they divide all the concentrations in half?

What is the equilibrium constant, K, at 400K if 3.00 moles of CO2(g) introduced into a 2 liter container are 20.0% dissociated at equilibrium?
2 CO2(g) « 2CO(g) + O2(g)




A. (0.30)^2(0.15) / (1.20)^2 mol/liter
B. (1.20)^2 / (0.30)^2(0.15) mol/liter
C. (0.60)^2(0.30) / (2.40)^2 mol/liter
D. (2.40)^2 / (0.60)^2(0.30) mol/liter
E. (0.30)^2(0.15) / (2.40)^2 mol/liter


I said the answer was C

 

[seoul]

You're dividing all moles (NOT the concentrations) with the container volume of 2 liter to yield the respective concentrations...

 

[Tracy47]

Thanks S!

I have another question. 🙂

What is the molality of 85% phosphoric acid, H3PO4, if the density of the solution is 1.70 g/ml?



A. 1000 * (1.70/1) * (100.0 / 15.0) * (1.70/1) * (85.0 / 100.0) * (1/98) m
B. 1000 * (1/1.70) * (100.0 / 15.0) * (1.70/1) * (85.0 / 100.0) * (1/98) m
C. 1000 * (1/1.70) * (100.0 / 15.0) * (1/1.70) * (85.0 / 100.0) * (1/98) m
D. 1000 * (1/1.70) * (15.0 / 100.0) * (1.70/1) * (85.0 / 100.0) * (1/98) m
E. 1000 * (1/1.70) * (100.0 / 15.0) * (1.70/1) * (100.0 / 85.0) * (1/98) m


The Answer is B, whihc I don't understand. The 100s get cancelled out, and the 1.70 get cancelled as well. what we're left with is 1000 * 85/(15*98), I don't particulary agree with that asnwer, as it doesn't give us the molality. Any ideas?

 

[Tracy47]

Ahh, DAT achiever is really confusing me.

It says:
∆H = -∑∆H(product bonds) + ∑∆H(reactant bonds)

but isn't it

∆H = ∑∆H(product bonds) - ∑∆H(reactant bonds)


They are obviously very different, and will give you different enthalpy change and will screw up whether the reaction is exo or endo.

Or am i missing something? This is #70 on Exam 2.

Thank you! 😉

 

[HansWalker]

What equations are you guys using to solve the first problem?

I'm lost, please help.

 

[seoul]

Thanks S!

I have another question. 🙂

What is the molality of 85% phosphoric acid, H3PO4, if the density of the solution is 1.70 g/ml?



A. 1000 * (1.70/1) * (100.0 / 15.0) * (1.70/1) * (85.0 / 100.0) * (1/98) m
B. 1000 * (1/1.70) * (100.0 / 15.0) * (1.70/1) * (85.0 / 100.0) * (1/98) m
C. 1000 * (1/1.70) * (100.0 / 15.0) * (1/1.70) * (85.0 / 100.0) * (1/98) m
D. 1000 * (1/1.70) * (15.0 / 100.0) * (1.70/1) * (85.0 / 100.0) * (1/98) m
E. 1000 * (1/1.70) * (100.0 / 15.0) * (1.70/1) * (100.0 / 85.0) * (1/98) m


The Answer is B, whihc I don't understand. The 100s get cancelled out, and the 1.70 get cancelled as well. what we're left with is 1000 * 85/(15*98), I don't particulary agree with that asnwer, as it doesn't give us the molality. Any ideas?

Click on th following for added explanation by dlink:

http://forums.studentdoctor.net/showthread.php?t=225090

 

[seoul]

Ahh, DAT achiever is really confusing me.

It says:
∆H = -∑∆H(product bonds) + ∑∆H(reactant bonds)

but isn't it

∆H = ∑∆H(product bonds) - ∑∆H(reactant bonds)


They are obviously very different, and will give you different enthalpy change and will screw up whether the reaction is exo or endo.

Or am i missing something? This is #70 on Exam 2.

Thank you! 😉

Tracy, don't confuse yourself. we're talking here about bond energies (NOT Enthalpies of Formation)...

Click on the following link that had the same question puzzling Denro too and pay attention to how the problem was clarified:

http://forums.studentdoctor.net/showthread.php?t=223691

 

[Tracy47]

seoul,

so thats the formula for bond energy? i thought we're using the bond energy to find the enthalpy change, and in this case, we're using enthalpy to find the bond energy. We should be able to use the same equation which is sum product minus sum reactants which equals enthalpy change.

 

[seoul]

seoul,

so thats the formula for bond energy? i thought we're using the bond energy to find the enthalpy change, and in this case, we're using enthalpy to find the bond energy. We should be able to use the same equation which is sum product minus sum reactants which equals enthalpy change.

I just edited the post... click on the above link if possible.

 

[Tracy47]

So bond energies and heat of formations are different? Aiya, I feel silly now =) 😛

 

[joonkimdds]

Geeezz...i learned it about a year ago and now i forgot everything that I don't even know what u r asking... 😱 I wonder how i can prepare for my DAT exam in 1 year....