HW 8.67  [ENDORSED]

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Christopher Reed 1H
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HW 8.67

Postby Christopher Reed 1H » Sun Jan 15, 2017 6:33 pm

Hi all.

I am stuck on letter a) of number 67 of the chapter 8 homework.

Looking at the solution manual, I understand the equation and all the calculations that lead to the change in enthalpy being -242kJ/mol. However, I think I am misunderstanding what this value represents. Why is it that -242kJ/mol is used "to produce water in the gas phase?" I don't understand why we need to take into account the enthalpy of vaporization after calculating -242kJ/mol.

If someone could walk me through what to do after calculating the change in enthalpy from the breaking and forming of bonds, I would greatly appreciate it.

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Re: HW 8.67

Postby Chem_Mod » Sun Jan 15, 2017 6:54 pm

Can you please adhere to the forum guidelines and post the question in whole so that we may all understand what you are asking about?

Christopher Reed 1H
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Re: HW 8.67

Postby Christopher Reed 1H » Sun Jan 15, 2017 10:47 pm

Sorry about the confusion, let me state the question in its entirety.

The question states:

"Use the information in Tables 8.3, 8.6, and 8.7 to estimate the enthalpy of formation of each of the following compounds in the liquid state."

I am specifically interested in how to do this for H2O.

The values of interest from the tables mentioned in the problem are 436kJ/mol to break H-H bonds, 496 kJ/mol to break O=O bond, and -463 KJ/mol to form O-H bond.

Using the above values and the chemical equation H2(g) + 1/2O2(g) --> H2O(l) I calculated the change in enthalpy to be -242kJ/mol. At this point the solution subtracts the enthalpy of vaporization of water (44.0 kJ/mol) to obtain the final answer of -286 kJ/mol.

My questions are as follows:
1. Why must I subtract the enthalpy of vaporization from the change in enthalpy?
2. If the question asked me to estimate the enthalpy of formation in the SOLID state, how would my calculations differ?

Vincent Tse 2B
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Re: HW 8.67  [ENDORSED]

Postby Vincent Tse 2B » Sun Jan 15, 2017 11:21 pm

When you consider water, for example, you have to have an idea of what your desire equation should be..

Since we're working with standard reaction enthalpy and standard enthalpy of formation, we have to remember that the formation of a compound must be created from its elements in their most stable state.

So: H2 + O2 --> H2O

Now, you balance the equation. For these problems, I noticed that fractional coefficients are common so it's okay to have them.

Balance rxn: H2 + 1/2 O2 --> H2O

*Remember that the elements are in their gaseous state. When calculating the standard enthalpy of formation for H2O, you're actually get a value that will produce water in the gas phase.

Recall that, liquid --> vapor is a vaporization process and has an associated enthalpy of vaporization. The following equation is ΔHvap = Hvapor - Hliquid

Instead, we just solved for the standard enthalpy of formation for water in its vapor form. Therefore, we know Hvapor.

ΔHvap can be obtained from a given table; thus, we rearrange and solve for Hliquid. This is the standard enthalpy of formation of water in its liquid state. And, so, this is the reason for why you subtract the enthalpy of vaporization from the standard enthalpy of formation of water in its vapor form.

204751840 2E
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Re: HW 8.67

Postby 204751840 2E » Wed Jan 18, 2017 10:49 pm

I have a question regarding b)

The question is finding the enthalpy of formation for the liquid state of CH3O3, methanol. According to the solutions manual, you have to add the enthalpy of sublimation for carbon, because carbon is a solid in its ground state. Is there any reason why we use the enthalpy of sublimation as opposed to the enthalpy of fusion and then the enthalpy of vaporization of carbon? Is carbon unable to exist in a liquid state? Thanks in advance.

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Re: HW 8.67

Postby Chem_Mod » Thu Jan 19, 2017 8:14 pm

Hello! The enthalpy of vaporization added to the enthalpy of fusion is the enthalpy of sublimation


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