1) For a reaction delta Go is more negative than delta Ho. What does this mean?

a. delta So is zero.
b. delta So is negative
c. delta So is positive
d. delta So is negative if delta Ho is positive.
e. delta So is negative if delta Ho is negative.

I think that delta S is negative if delta H is positive.

2) Sodium reacts violently with water according to the equation:

2 Na (s) + 2 H2O (l) -> 2 NaOH (aq) + H2 (g)

The resulting solution has a higher temperature than the water prior to the addition of the sodium. What are the signs of delta Ho and delta So for this reaction?

a. delta Ho is negative and delta So is positive.
b. delta Ho is positive and delta So is negative.
c. delta Ho and delta So are both negative.
d. delta Ho and delta So are both positive.

I think that the water is endothermic which would make the system exothermic which means that delta H and delta S would be negative.

Would you agree with my answers? Thank you.

1. Let's put some numbers in.

dG = dH - TdS
-200 = -100 - TdS.
dG is more negative than dH.
The only way this can work is for dS to be +. If dS is - then -TdS is + and -100 + any number means dG is not more negative than dH.

+90 = +100 - TdS
dG is still more negative than dH.
If dS is +, then -TdS is - and any - number added 100 makes it smaller so dG < dH.
If dS is -, then - TdS is + and any + number added to 100 makes dG more positive than dH and not more negative than dH.
I think the only answer is c.

2. I agree that water is absorbing heat and that the reaction is exothermic. That means dH must be -. Look at the reaction.
2Na(s) + 2H2O(l) ==> H2(g) + 2NaOH(aq)
solid to aqueous. liquid to gas. Doesn't that spell + for S?

For question 2 my actually chose that delta H is negative and delta s is positive I just typed the wrong answer in. Is this correct?

Yes, a is correct.

For the first question, you correctly identified that if ΔG° (standard Gibbs free energy change) is more negative than ΔH° (standard enthalpy change), it means that ΔS° (standard entropy change) is negative. So option d. "ΔS° is negative if ΔH° is positive" is the correct answer.

For the second question, you stated that the water is endothermic which would make the system exothermic and both ΔH° and ΔS° negative. However, this is not accurate. The reaction given is the reaction of sodium with water, which is highly exothermic and releases large amounts of heat. Therefore, the correct answer is a. "ΔH° is negative and ΔS° is positive."

To determine the signs of ΔH° and ΔS°, you can use the reaction equation and consider the nature of the reaction. In this case, since the resulting solution has a higher temperature than the water prior to the addition of sodium, it indicates that heat is being released, indicating a negative value of ΔH°. Moreover, the reaction involves the creation of gas (hydrogen) and the formation of an aqueous solution (NaOH), suggesting an increase in entropy, resulting in a positive value of ΔS°.

So, while you were correct in the second part, you have one incorrect answer in the first part.