Hess' Law


by Greyt
Tags: hess
Greyt
Greyt is offline
#1
Feb2-13, 05:13 PM
P: 4
Hello there,

I am supposed to experimentally determine (with the use of a calorimeter) the overall heat of reaction per mole of calcium metal in two cases:

1. When I mix HCL, calcium and water in that order.
2. When I mix water, calcium and HCL in that order.

In each case, I am supposed to treat the overall reaction as a single process and determine a ΔH for the two-step process using my determined overall change in temperature.

I've purposefully left out the provided numerical values, hoping to understand more where to start than to solve the actual problem at the moment; however I can provide them if desired.

From what I understand, I'm supposed to basically sum up the overall reaction in an equation and that both ways should theoretically result in the release of the same amount of energy... but how on earth would I go about calculating the heat of reaction using the overall temperature change?

I will mention that we are provided with the following:

1. Mass of the solution
2. Initial and final temperature
3. Mass of each reagent and the calorimeter itself
4. Heat capacity of the calorimeter

Depressing as it is, I'm fairly lost and any information would be appreciated, thanks.

NOTE: My current attempt revolves around treating the solution as having a heat capacity similar to water, and having the ΔH = CmΔT(water) + CΔT (calorimeter). I don`t know if it`s very appropriate to assume such a thing however and would like to see if there`s any other ideas.
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Mandelbroth
Mandelbroth is offline
#2
Feb2-13, 06:33 PM
Mandelbroth's Avatar
P: 597
Quote Quote by Greyt View Post
Hello there,

I am supposed to experimentally determine (with the use of a calorimeter) the overall heat of reaction per mole of calcium metal in two cases:

1. When I mix HCL, calcium and water in that order.
2. When I mix water, calcium and HCL in that order.

In each case, I am supposed to treat the overall reaction as a single process and determine a ΔH for the two-step process using my determined overall change in temperature.

I've purposefully left out the provided numerical values, hoping to understand more where to start than to solve the actual problem at the moment; however I can provide them if desired.

From what I understand, I'm supposed to basically sum up the overall reaction in an equation and that both ways should theoretically result in the release of the same amount of energy... but how on earth would I go about calculating the heat of reaction using the overall temperature change?

I will mention that we are provided with the following:

1. Mass of the solution
2. Initial and final temperature
3. Mass of each reagent and the calorimeter itself
4. Heat capacity of the calorimeter

Depressing as it is, I'm fairly lost and any information would be appreciated, thanks.

NOTE: My current attempt revolves around treating the solution as having a heat capacity similar to water, and having the ΔH = CmΔT(water) + CΔT (calorimeter). I don`t know if it`s very appropriate to assume such a thing however and would like to see if there`s any other ideas.
Important to remember in this case is that the reaction is taking place at what we can assume to be constant pressure. In this case, for which we chemists use the fancy term "isobaric", the heat of the reaction is equal to the enthalpy of the reaction. Thus, you are on the right track with your above attempt.

We can assume that the heat capacity is similar to water if your solution is dilute enough, though you have not given us enough information to confirm that. However, your change in enthalpy should be for the whole solution. Thus, we can just write [itex]\Delta H = Q_{reaction} = (c_{H_{2}O})(m_{sol'n})(\Delta T_{sol'n})[/itex].
Greyt
Greyt is offline
#3
Feb2-13, 07:12 PM
P: 4
Quote Quote by Mandelbroth View Post
Important to remember in this case is that the reaction is taking place at what we can assume to be constant pressure. In this case, for which we chemists use the fancy term "isobaric", the heat of the reaction is equal to the enthalpy of the reaction. Thus, you are on the right track with your above attempt.

We can assume that the heat capacity is similar to water if your solution is dilute enough, though you have not given us enough information to confirm that. However, your change in enthalpy should be for the whole solution. Thus, we can just write [itex]\Delta H = Q_{reaction} = (c_{H_{2}O})(m_{sol'n})(\Delta T_{sol'n})[/itex].
First off, thank you for the swift reply.

It seems that is the correct method of solving the problem, but I am curious as to when I can use the heat capacity of water. Is there a general guideline as to how concentrated a solution has to be before you can no longer say its heat capacity is similar to that of water? At the moment, I seem to rely on that method unless I am dealing with a solution that's stated to be nearly void of water.

Mandelbroth
Mandelbroth is offline
#4
Feb3-13, 02:57 PM
Mandelbroth's Avatar
P: 597

Hess' Law


Quote Quote by Greyt View Post
First off, thank you for the swift reply.

It seems that is the correct method of solving the problem, but I am curious as to when I can use the heat capacity of water. Is there a general guideline as to how concentrated a solution has to be before you can no longer say its heat capacity is similar to that of water? At the moment, I seem to rely on that method unless I am dealing with a solution that's stated to be nearly void of water.
Heat capacity is the amount of heat (energy transfer) required to change the temperature of a substance by some amount. We are using specific heat capacity, which is the heat capacity per unit of mass.

Because our "heat capacity" depends on mass, we can infer that it changes with the increase of solute (more solute implies more mass). You will likely study the colligative properties of solutions. By adding more solute to a solution, you are reducing the chemical potential, which both lowers the freezing point and raises the boiling point. The concept is rooted in this. However, I think that the change in your accuracy by using the specific heat of water rather than a more exact value for your solution will not be a problem for now.


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