General Chemistry.......trick question

[Toothguy80]

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150 mL of a 3M KCL solution is diluted with 500 mL of distilled water. The resulting solution is stirred and 300 mL are poured out of the beaker. What is the molarity (in moles/liter) of the remaining solution?

The following equilibrium takes place:
2A(g) + 3B(g) --> 4C(g) + 2D(g) DeltaH= -50kcal/mol
What set of conditions will cause increased formation of C?
a)Increased Temperature
b)Decreased Temperature
c)Increased Pressure
d)Decreased Volume
e)None of the above

I thought that the answer was a) because when volume increases, moles and temp. increase too, right? and Pressure decreases, right?
But the answer is b) because the reaction is exothermic. But is my reasoning wrong? If they didn't give the DeltaH of rxn, then would I be right?

 

[shane.]

The following equilibrium takes place:
2A(g) + 3B(g) --> 4C(g) + 2D(g) DeltaH= -50kcal/mol
What set of conditions will cause increased formation of C?
a)Increased Temperature
b)Decreased Temperature
c)Increased Pressure
d)Decreased Volume
e)None of the above

I thought that the answer was a) because when volume increases, moles and temp. increase too, right? and Pressure decreases, right?
But the answer is b) because the reaction is exothermic. But is my reasoning wrong? If they didn't give the DeltaH of rxn, then would I be right?

No, you'd still be wrong..

With an increase in pressure a system will shift the equilibrium in a manner such as to decrease the number of moles of gas present. In other words, the equilibrium will shift to the side of the reaction with the least number of moles of gas and produce more of those/that species. Decreased volume = increased pressure remember...

I think you're misreading option a) because if you understand that b) is the correct choice there is no way you would have chosen a)...

 

[Anemone]

decreased temp, rxn goes exotherm
increased temp,rnx goes endotherm

 

[Streetwolf]

Just imagine heat being either a reactant or a product. If delta H is negative then it's exothermic and heat is a product. If delta H is positive then it's endothermic and heat is a reactant. Then you know what happens when a product is removed/added or a reactant is removed/added... But instead of heat being 'removed/added' the temperature is increased (heat added) or decreased (heat removed).

 

[Sillygirl]

Is the first answer 0.7 M?

 

[pluginchugin]

is the first answer 3.0 M?

 

[shane.]

I get 0.7M as well.

 

[Dotoday]

Yes, it is 0.7 M because the pouring of 300ml out of the beaker makes no difference since the solution is already dissolved.

 

[Toothguy80]

No, you're all wrong...see, this is such a trick question.

The answer is none of the above because it doesnt change the moles only because water is added.

If you add water, you change the volume, but not the moles....

 

[shane.]

No, you're all wrong...see, this is such a trick question.

The answer is none of the above because it doesnt change the moles only because water is added.

If you add water, you change the volume, but not the moles....

Good thing Toothguy is here to catch our mistakes...

Read your response, then read the question again.

 

[marianirj]

150 mL of a 3M KCL solution is diluted with 500 mL of distilled water. The resulting solution is stirred and 300 mL are poured out of the beaker. What is the molarity (in moles/liter) of the remaining solution?

The following equilibrium takes place:
2A(g) + 3B(g) --> 4C(g) + 2D(g) DeltaH= -50kcal/mol
What set of conditions will cause increased formation of C?
a)Increased Temperature
b)Decreased Temperature
c)Increased Pressure
d)Decreased Volume
e)None of the above

I thought that the answer was a) because when volume increases, moles and temp. increase too, right? and Pressure decreases, right?
But the answer is b) because the reaction is exothermic. But is my reasoning wrong? If they didn't give the DeltaH of rxn, then would I be right?

Isn't the first problem M1V1= M2V2. So M1 = 3, V1 = .65 L (.15 KOH and .5 H2O), and V2 = .35 (.65 - .3). Therefore the final molarity is (3 x .65) / .35 = ~5.5 ????

And isn't the answer to the second problem b because it is an exothermic reaction because H is negative. Since the energy is released, increasing temp would cause reaction to go left, and decreasing it would go right?

 

[schmick]

The answer is none of the above because it doesnt change the moles only because water is added.

If you add water, you change the volume, but not the moles....

You are right, the number of moles does not change but the molarity DOES change as the volume changes.

It's a simple M1V1=M2V2 question:
3M x 150ml = Z x 650ml
Z = 0.69M

It doesn't matter that you pour 300ml out of the final soulution because you are also losing some of the KCl.

 

[schmick]

And isn't the answer to the second problem b because it is an exothermic reaction because H is negative. Since the energy is released, increasing temp would cause reaction to go left, and decreasing it would go right?

The answer is B. It is very helpful to think of heat in exo/endothermic rxns as a reactant or product. Somebody has explained it very well above.

 

[marianirj]

You are right, the number of moles does not change but the molarity DOES change as the volume changes.

It's a simple M1V1=M2V2 question:
3M x 150ml = Z x 650ml
Z = 0.69M

It doesn't matter that you pour 300ml out of the final soulution because you are also losing some of the KCl.

I agree, I think the answer is .69.

 

[mhamilton23]

I got 1.38M and 0.69M, depending if you want the molarity of all ions (K+ and Cl-, not including H+ and OH- from the dissociation of water) or the molarity of one of the ions.

1. (3M)(0.15L) = 0.45 moles of K+ before dilution (same for Cl-)
2. 0.45 moles/0.65L (new volume after dilution) = 0.692M for K+ after dilution
3. Removing 300 mL of HOMOGENOUS solution will NOT change the concentration (although it will change the # of moles). For argument sake, we can calculate this to make sure:
(0.692M)(.300L) = 0.208 moles of K+ lost in the 300 ml that was dumped. Therefore, we have (0.45 moles - 0.208 moles) 0.242 moels of K+ left in (150ml+500ml-300ml) 350 mL of solution left over, which yields a final concentration of (0.242mol/0.35L) 0.69M K+ (or Cl-). Doubling this will give the concentration of ALL ions in the solution.

I still fail to see how we are wrong, and how this is a "trick" question. But please enlighten us if we are truly incorrect.

Thanks

 

[Dotoday]

Yes I agree with schmick andmhamilton23 , the answer is 0.69. You cannot Not change the molarity of salt after adding a large volume of water to it. KCL is soluble and it says that it is stored, so it must therefore dilute the salt. I believe there is a mistake in the book somewhere. Also since solution becomes homogenous (it must) therefore poring any volume out of it will not change the total molarity of the solution.

 

[Pebbles000]

what book are these problems coming from????


I got .69 as well btw.