Give me some responses to the questions below and I'll give you some extra credit.

SID#:

Laboratory Section: 1 2 3 4 5 6

1. The enthalpy of formation, H^o_f , for an unknown compound, AB, in the solid phase and the gaseous phase were determined by experiment. The person doing the experiment forgot to associate the value of the H^o_f to the specific phase. Can you help? The two values were; H^o_f = -23 kJ and -35 kJ. Explain your reasoning.

H^o_f = -35 kJ is most likely the enthalpy of formation for the solid.

H^o_f = -23 kJ is most likely the enthalpy of formation for the gas.

The reason the numbers are like this is because it requires heat to form the gas phase from the solid, that energy must be absorbed by the gas. Another way it to think that solids are lower in energy.

Finally remember the conversion of the solid to the vapor must be endothermic.

AB(s) ---> AB(g)

H = H^o_f (P) – H^o_f (R)

H = -23 kJ - (-35 kJ) = +12 kJ

2. Given a balanced chemical equation explain how you would determine the H^o_rxn?

3. Briefly explain what the term entropy means to you.

Entropy is a measure of the dissorder, or molecular randomness, of a system. The greater the dissorder or randomness of a system the higher its entropy.

4. Identify the phase of the substance in each of the containers labeled A, B and C. Indicate the phase that is most ordered, and most disordered.

A is a solid, B is a liquid and C is a gas. The solid is the most order physical state, liquids are the next most order, and gases are the least ordered (most dissordered).

5. Indicate for each change if the products are more dissordered or less disorderd (more ordered) compared to the reactants.

a. H₂O(l) ----> H₂O(s) products are more ordered compared to the reactants

b. CO₂(s) ----> CO₂(g) products are more dissordered compared to the reactants

c. 2Na(s) + 2H₂O(l) ----> 2NaOH(aq) + H₂(g) products are more dissordered compared to the reactants

d. 2H₂(g) + O₂(g) ----> 2H₂O(g) products are more ordered compared to the reactants

6. When the products are more dissordered compared to the reactants we say S for the reaction is positive, and if the products are more ordered compared to the reactants we say the S for the reaction is negative. For the following reactions predict the sign of S.

a. 2H₂(g) + O₂(g) ----> 2H₂O(g) S is -

b. Ba(OH)₂^.8H₂O(s) + 2NH₄SCN(s) ----> Ba(SCN)₂(aq) + 10H₂O(l) + 2NH₃(aq) S is +

c. 2NO(g) + 2CO(g) ----> N₂(g) + 2CO₂(g) S is -

d. C₃H₈(g) + 5O₂(g) ----> 3CO₂(g) + 4H₂O(l) S is -

7. In each of the following pairs which is more dissordered.

a) H₂O(l) or H₂O(g) H₂O(g) is more dissordered

b) H₂O(l) at 25 degrees Celsius or H₂O(l) at 90 degrees Celsius H₂O(g) at 90 degrees Celsius is more dissordered

8. Is there anything about the questions that you feel you do not understand? List your concerns/questions.