TBR Thermochemistry

[sacha]

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In TBR, eg 8.22 (pg 154)

Calorimetry experiment; mixing acid & base in a styrafoam cup

what would be observed if concentrations of HCl & KOH were cut in half?
TBR ans: change in temperature would be cut in half

BR explanation: if concentration of acid & base solutions are cut in half, means moles of reactants are cut in half & amount of heat released is cut in half. this would reduce change in temperature by half.

but i'm confused about the link between heat released & temperature change. according to q=mc(change in T), wouldn't moles being cut in half mean both m & q is cut in 1/2 & temperature change should remain the same??

 

[ridethecliche]

In TBR, eg 8.22 (pg 154)

Calorimetry experiment; mixing acid & base in a styrafoam cup

what would be observed if concentrations of HCl & KOH were cut in half?
TBR ans: change in temperature would be cut in half

BR explanation: if concentration of acid & base solutions are cut in half, means moles of reactants are cut in half & amount of heat released is cut in half. this would reduce change in temperature by half.

but i'm confused about the link between heat released & temperature change. according to q=mc(change in T), wouldn't moles being cut in half mean both m & q is cut in 1/2 & temperature change should remain the same??

Intuitively, if you have the same amount of solution, but the molarity is halved, then you're going to generate half the amount of heat. Half the amount of heat is going to lead to half the rise in temperature.

M is a measure of the mass of the solution. This doesn't change between the trials. I mean, I guess it does because of the solutes, but you're just going to end up with salt in the solution and water due to neutralization.

Does that make sense? q is the only thing that's really changing.

 

[sacha]

Intuitively, if you have the same amount of solution, but the molarity is halved, then you're going to generate half the amount of heat. Half the amount of heat is going to lead to half the rise in temperature.

M is a measure of the mass of the solution. This doesn't change between the trials. I mean, I guess it does because of the solutes, but you're just going to end up with salt in the solution and water due to neutralization.

Does that make sense? q is the only thing that's really changing.

hey🙂 yeah, i get it now. i just assumed half the moles would weigh half as much but now I see the mistake in my train of thought, thanks!

 

[MedChallenge]

Intuitively, if you have the same amount of solution, but the molarity is halved, then you're going to generate half the amount of heat. Half the amount of heat is going to lead to half the rise in temperature.

M is a measure of the mass of the solution. This doesn't change between the trials. I mean, I guess it does because of the solutes, but you're just going to end up with salt in the solution and water due to neutralization.

Does that make sense? q is the only thing that's really changing.

Hey, I don't have access to this question so would you please explain this again? Why do you say the mass (or anything else, but q) doesn't really change?

 

[deleted393595]

Hey, I don't have access to this question so would you please explain this again? Why do you say the mass (or anything else, but q) doesn't really change?

The Q=mCT approach is incorrect in this situation. This equation is used to measure how much heat a material absorbs/releases and not for reactions. The formula you're looking for is:

Heat of Formation = Joules/mole

If you cut down your molar concentration by half, you're cutting down the number of moles by half. Thus, you get half the released heat.

 

[Rabolisk]

Well, Q = mcΔT isn't wrong. In fact you need it because the question asked for the final effect on the temperature. In a solution, though, halving the concentration doesn't decrease the mass of the solution. The mass of the solution is more or less entirely attributable to the solvent.

To be nitpicky, you can argue that you can cut concentration in half by doubling the volume, which results in double the mass, which results in same temperature change. But that's clearly not what they mean.

 

[MedChallenge]

Well, Q = mcΔT isn't wrong. In fact you need it because the question asked for the final effect on the temperature. In a solution, though, halving the concentration doesn't decrease the mass of the solution. The mass of the solution is more or less entirely attributable to the solvent.

To be nitpicky, you can argue that you can cut concentration in half by doubling the volume, which results in double the mass, which results in same temperature change. But that's clearly not what they mean.

You must have killed the MCAT. Thanks, again.

 

[Rabolisk]

Looking over it again, I misspoke. The temperature change wouldn't be the same, but would be 1/4th the original. Not that it matters.

 

[ridethecliche]

Well, Q = mcΔT isn't wrong. In fact you need it because the question asked for the final effect on the temperature. In a solution, though, halving the concentration doesn't decrease the mass of the solution. The mass of the solution is more or less entirely attributable to the solvent.

To be nitpicky, you can argue that you can cut concentration in half by doubling the volume, which results in double the mass, which results in same temperature change. But that's clearly not what they mean.

Eggsactly!

I think it's easier to think about it that way for this particular problem.

I think the red part is wrong though. Different temperature change since you have 2x the solvent, but same amount of heat change.

Edit: Just saw your correction

 

[BerkReviewTeach]

Looking over it again, I misspoke. The temperature change wouldn't be the same, but would be 1/4th the original. Not that it matters.

You might want to double check this by putting numbers in.

Let's say you cut the concentration in half by doubling the volume, as you suggest. If you originally have 100 mL 1.0 M HCl and 100 mL 1.0 M NaOH and you add 100 mL of water to both solutions, then you have 200 mL 0.5 M HCl and 200 mL 0.5 M NaOH.

The moles of acid and moles of base have not changed with the dilution, so you'll get the same amount of heat from the reaaction (no change in q). But the final mass of the solution is now doubled (400 mL rather than 200 mL), so according to q = mCdeltaT, if q remains the same, C is essentially the same, and m is doubled, then deltaT must be half as large as before.

The temperature change should be half as great when the concentrations are cut in half, no matter what volumes you mix, as long as you're mixing equal volumes of acid as base.

 

[Rabolisk]

That's right. My apologies. I mistakenly thought that heat of reaction is dependent on concentration, rather than moles.

 

[BerkReviewTeach]

That's right. My apologies. I mistakenly thought that heat of reaction is dependent on concentration, rather than moles.

If I'm counting correctly, a minor error here makes you about 581/582. You are a friggen genius who clears up so many threads around here with clean, logical solutions. You are a godsend to this forum. I think an occassional brainfart (once every 582 questions) might be acceptable. There have been so many threads I clicked on to add my two cents when after reading your response I realized that there was nothing to add. You give clear, well-thought solutions that get to the crux of the question. Thanks for all you do for the Q&A forum.

 

[Catburr]

You give clear, well-thought solutions that get to the crux of the question. Thanks for all you do for the Q&A forum.

I'll second that. Thanks a bunch Rabo, you definitely have a knack for clear, thorough explanations of an astoundingly wide variety of problems. If I manage a solid score, I definitely credit Rabo, BerkTeach, and many of the other positive contributors here as having no small part in that!

 

[sacha]

You might want to double check this by putting numbers in.

Let's say you cut the concentration in half by doubling the volume, as you suggest. If you originally have 100 mL 1.0 M HCl and 100 mL 1.0 M NaOH and you add 100 mL of water to both solutions, then you have 200 mL 0.5 M HCl and 200 mL 0.5 M NaOH.

The moles of acid and moles of base have not changed with the dilution, so you'll get the same amount of heat from the reaaction (no change in q). But the final mass of the solution is now doubled (400 mL rather than 200 mL), so according to q = mCdeltaT, if q remains the same, C is essentially the same, and m is doubled, then deltaT must be half as large as before.

The temperature change should be half as great when the concentrations are cut in half, no matter what volumes you mix, as long as you're mixing equal volumes of acid as base.

that's a really clear explanation. I was also under the impression that q is based on concentration instead of moles. thanks for clearing that up.

 

[DPTinthemaking15]

Sorry for necrobumping this, but I am having the hardest time with this problem. Should you use Q=mct, if so, how does concentration fit into this equation? To be honest, I assumed that only temperature, mass, and the specific heat could fit into this equation, not concentration. I answered the problem correct through reasoning, but I wanted to understand it more in-depth in case a harder problem were to arise.