Question 8.25

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K Stefanescu 2I
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Joined: Fri Sep 29, 2017 7:04 am

Question 8.25

Postby K Stefanescu 2I » Mon Jan 08, 2018 3:47 pm

Quesiton:
A constant-volume calorimeter was calibrated by carrying out a reaction known to release 3.50 kJ of heat in 0.200 L of solution in the calorimeter (q=-3.50 kJ), resulting in a temperature rise of 7.32 degrees C. In a subsequent experiment, 100.0 mL of 0.200 m HBr(aq) and 100.0 mL of 0.200 m KOH(aq) were mixed in the same calorimeter and the temperature rose by 2.49 degrees C. What is the change in the internal energy of the neutralization reaction?

I understood that we need to calculate the C of the calorimeter. However, I do not understand how to solve the rest of the problem. Why does the solutions manual solve it as it does, by seemingly ignoring the information regarding how much of each reactant was present? Thank you in advance.

Yashaswi Dis 1K
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Joined: Fri Sep 29, 2017 7:04 am

Re: Question 8.25

Postby Yashaswi Dis 1K » Mon Jan 08, 2018 8:53 pm

I think the value that we really need to find here is change in internal energy (delta U).

We know that delta U is equal to q + w. Since the question involves a bomb-calorimeter (constant volume), we know that w should equal zero since w=-P*deltaV and delta V is zero. So we only need to really find the value q of the reaction.

After you find out the value of Cv using delta U/delta T (7.32 degrees celsius), you have to find qcal (heat of the calorimeter). qcal = Cv * Temperature(2.49 degrees celsius).

Since the heat is absorbed by the solution in the calorimeter, that means the same amount of heat is released by the reaction. So -qcal = q (for the reaction). There is your answer since delta U will be equal to q.

I think the content for mL and mol is just extra stuff that doesn't really help with finding the answer.

I hope this helps!

Sohini Halder 1G
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Re: Question 8.25

Postby Sohini Halder 1G » Mon Jan 08, 2018 9:40 pm

In this case, you can ignore the moles of reactant and product because those do not matter in the equation for finding q.

AtreyiMitra2L
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Re: Question 8.25

Postby AtreyiMitra2L » Mon Jan 08, 2018 10:43 pm

The rest of the question is not important. The question is asking us only to find the change in internal energy. Therefore, we should only be looking for changes in q and w. We find c (the heat capacity) so that we can use this value later to multiply it by the temperture change in the second experiment. This will give us q. We know that w will be 0 because the question explicitely states that its a constant volume calorimeter.

Humza_Khan_2J
Posts: 56
Joined: Thu Jul 13, 2017 3:00 am

Re: Question 8.25

Postby Humza_Khan_2J » Tue Jan 09, 2018 9:06 pm

This question wholeheartedly depends on q. W is 0 because dV is 0. Therefore, it should be easy using basic qcal calculations to figure this out!


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