Re: **Important Midterm Information Winter 2015**

[Chem_Mod]

I have uploaded my review session.
To view in a larger window click play and then right click on the video image, scroll to Zoom and select Full Screen.

I discussed how to solve these problems:

Q1.
When eating large amounts of candy, people can develop a “sugar rush.”
A typical sugar found in candy is glucose which will react with oxygen to form carbon dioxide and water.
Suppose on Halloween a child weighing 35 kg (which is mostly water) eats 2 butterfinger candy bars each containing 29.0 g of glucose. If all of the glucose reacted with oxygen inside the body and the heat evolved was used solely to raise the child’s body temperature from the standard 37oC, how hot would the child’s fever be? Assume the specific heat capacity of the child’s body is close to that of water.
Glucose molecular formula, C6H12O6 MW, glucose = 180.16 g.mol-1
∆Hof C6H12O6(s) = -1007 kJ.mol-1
∆Hof CO2(g) = -393.5 kJ.mol-1 ∆Hof H2O(l) = -285.83 kJ.mol-1

Q2.
During your camping weekend at the sea you used 2.00 moles of butane (C4H10) gas for cooking. The enthalpy of combustion of butane is -2878. kJ.mol-1. Assume the reaction occurs at 300. oC and a constant pressure of 1.00 atm. (a) Write the balanced reaction for combustion of butane gas. (b) Calculate the work associated with this combustion reaction. (c) Calculate the change in internal energy for the system.

Q3.
Calculate the temperature at which the decomposition of acetic acid to methane and carbon dioxide becomes spontaneous. Assume that ΔH° and ΔS° do not vary with temperature.
CH3COOH(l) → CH4(g) + CO2(g)

So(298K) J K-1 mol-1 ΔHfo(298K) kJ mol-1
CH3COOH(l) 158.8 -487.0
CH4(g) 186.3 -74.8
CO2(g) 213.6 -393.5

Q4.
Hydrofluoric acid, HF, is a weak acid. After dissolving 10.0 g of HF in 100 mL of H2O at room temperature the pH is determined to be 1.20. Hint: pH = -log [H+] HF, MW = 20.0 g.mol-1
A. Calculate ∆G°r for the dissociation reaction: HF(aq) ⇌ H+(aq) + F-(aq)
B. In an exact replication of the first experiment the pH is measured to 5.34 before the reaction reaches equilibrium. What is ∆Gr for this reaction?

Q5.
Given the following information, calculate the standard enthalpy for the following reaction:
2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(l)

∆G°f (H2O (l )) = -237.13 kJ.mol-1 S° (H2O (l)) = 69.91 J. K-1mol-1
∆G°f (C8H18 (l )) = + 6.4 kJ.mol-1 S° (C8H18 (l)) = 358.0 J. K-1mol-1
∆G°f (CO2 (g)) = -394.36 kJ.mol-1 S° (CO2 (g)) = 213.74 J. K-1mol-1
S° (O2 (g)) = 205.14 J. K-1mol-1
Assume room temp.

After viewing my review session if you have content related questions post them here and I will answer them.

[Chem_Mod]

I have uploaded my midterm review notes. Sorry it is a bit messy!
-Jason Fang

[Chem_Mod]

All Chem 14A and 14B quizzes and exams have a cover page with constants and equations. See your course reader and workbook for examples. If needed a periodic table is also given.

[Astred 1D]

This question is in regards to Professor Lavell's Review session problems:

During Professor Lavell's review session, problem number two (the camping at the sea problem) when the combustion reaction of butane was written the water was a gas. A problem similar to this in the text book, 7.105, wrote the reaction of C6H6(L) with carbon dioxide gas and water liquid as the products.
I spoke to a T.A. about the products in a combustion reaction and was told they are always CO2 (g) and H20(l)
Do we use water liquid or water gas, for combustion reactions?

Thank you

[Chem_Mod]

This question is in regards to Professor Lavell's Review session problems:

During Professor Lavell's review session, problem number two (the camping at the sea problem) when the combustion reaction of butane was written the water was a gas. A problem similar to this in the text book, 7.105, wrote the reaction of C6H6(L) with carbon dioxide gas and water liquid as the products.
I spoke to a T.A. about the products in a combustion reaction and was told they are always CO2 (g) and H20(l)
Do we use water liquid or water gas, for combustion reactions?

Thank you

In my review session Q2 states the reaction occurs at 300 ^oC and therefore H₂O is a gas.

If the reaction temperature is 25 ^oC then H₂O is a liquid.

If the reaction temperature is not given then assume it is 25 ^oC which is why H₂O is often a liquid.

Hope this clarifies.
Good luck to all with their midterm.

[RaquelAvalos1K]

To clarify for the butterfinger & child example, we are using the grams/mass of the child in the specific heat capacity equation because we are solving for the T_f of the what happens to the surroundings (child)?
Whereas we include the heat released since it is released into the surroundings, despite coming from the system?

Also, as far as the reaction being exothermic, why would there be a negative sign on both the left and right side of the heat capacity equation? Wouldn't it only be on one side, otherwise the signs would cancel to be positive?


Thank you!


(I had an EDIT for the first question, as I thought we were using the glucose mass amount)

[Chem_Mod]

To clarify for the butterfinger & child example, we are using the grams/mass of the child in the specific heat capacity equation because we are solving for the T_f of the what happens to the surroundings (child)?
Whereas we include the heat released since it is released into the surroundings, despite coming from the system?

Also, as far as the reaction being exothermic, why would there be a negative sign on both the left and right side of the heat capacity equation? Wouldn't it only be on one side, otherwise the signs would cancel to be positive?

Thank you!


(I had an EDIT for the first question, as I thought we were using the glucose mass amount)

The chemical reaction (glucose in the butterfinger candy bars) is the system (which gives off the heat).
The child is the surrounds (heat enters).

q(system) + q(surroundings) = 0
Which can be written:
q(system) = -q(surroundings)
or
q(surroundings) = -q(system) (this is what I used in the typed solution)
Both the above statements give the same answer.

q(surroundings) = -q(system) = - (-988 kJ) = 988 kJ