Educational worksheet on Gibbs Free Energy concepts for chemistry students.

Here are the answers to the questions on your worksheet about Gibbs Free Energy.

1. In your own words, explain what “spontaneous” means.
Spontaneous means that a process or reaction happens naturally on its own without needing any outside help or continuous energy input. Think of it like a ball rolling down a hill; once you let go, it rolls by itself. You don't have to push it.

2. In your own words, explain what “non-spontaneous” means.
Non-spontaneous means that a process will not happen on its own. It requires outside work or energy to make it happen. Using the same example, pushing a ball *up* a hill is non-spontaneous because you have to keep pushing (adding energy) to make it move that way.

3. Given the following equation circle the correct answer:
$\Delta G = \text{positive}$
$\Delta S = \text{negative}$

* What is $\Delta H$?
Since $\Delta G = \Delta H - T\Delta S$, and we know $\Delta G$ is positive ($+$) and $\Delta S$ is negative ($-$), the term $-T\Delta S$ becomes positive (because a negative times a negative is a positive).
So, $\Delta G (+) = \Delta H + (\text{positive number})$.
For the result to be positive, $\Delta H$ doesn't strictly *have* to be positive in all math cases, but in chemistry contexts for these specific "always/never" questions: If $\Delta H$ were negative, the reaction might become spontaneous at low temps. However, usually, if $\Delta S$ is negative (unfavorable) and $\Delta G$ is positive (non-spontaneous), it implies the enthalpy change $\Delta H$ is likely positive (endothermic/unfavorable) as well, making it non-spontaneous at all temperatures. Let's look at the options provided in similar questions below to see the pattern. The options usually ask about temperature dependence.
Actually, let's look at the logic:
$\Delta G > 0$
$\Delta S < 0 \rightarrow -T\Delta S > 0$
$\Delta G = \Delta H - T\Delta S$
Positive = $\Delta H$ + Positive Term.
This equation holds true if $\Delta H$ is positive. If $\Delta H$ is negative, it could still be positive overall if $T$ is large enough. But typically, this combination ($\Delta H > 0, \Delta S < 0$) leads to a non-spontaneous reaction at all temperatures.

*Let's re-read the specific question format.* It asks "What is $\Delta H$?" and gives options about temperature. Wait, looking at Question 4, 5, 6, they follow a pattern. Question 3 just asks "What is $\Delta H$?" with options:
a. always spontaneous...
b. always non-spontaneous...
c. negative at lower temps...
d. always negative

Actually, looking closely at the image text for #3:
"Given... $\Delta G = \text{positive}$, $\Delta S = \text{negative}$. What is $\Delta H$?"
The options listed under #3 seem to be describing the *spontaneity* or the sign of $\Delta H$? No, the options a-d describe spontaneity conditions. This is slightly confusingly labeled. Let's look at option C and D. They mention "negative at lower/higher temperatures". This usually refers to $\Delta G$.

Let's assume the question asks: Under what conditions is this reaction spontaneous/non-spontaneous? OR What is the sign of $\Delta H$?
If $\Delta G$ is explicitly stated as positive, the reaction is non-spontaneous.
If $\Delta S$ is negative, the entropy term opposes spontaneity.
For $\Delta G$ to be positive when $\Delta S$ is negative, $\Delta H$ is most likely positive.
Reaction type: Endothermic ($\Delta H > 0$) and Decreasing Entropy ($\Delta S < 0$).
Result: Non-spontaneous at ALL temperatures.

Looking at the options provided in the crop for #3:
a. always spontaneous at all temperatures
b. always non-spontaneous at all temperatures
c. negative at lower temperatures, positive at higher temperatures
d. always negative

The correct description for a reaction where $\Delta H > 0$ and $\Delta S < 0$ is that it is always non-spontaneous. Therefore, the answer corresponds to option b. (Note: The question asks "What is $\Delta H$?", but the options describe the reaction behavior. In many worksheets, this is a typo for "What is the spontaneity?" or the student is meant to identify the scenario. Given $\Delta G$ is already defined as positive, the reaction IS non-spontaneous. The only scenario that fits $\Delta G(+)$ and $\Delta S(-)$ consistently without temperature dependence flipping it is if $\Delta H$ is also positive. The result is always non-spontaneous.)

*Correction*: Let's look at the other questions to understand the template.
Q4: $\Delta G = \text{negative}, \Delta S = \text{positive}$. Options: spontaneous/non-spontaneous/temp dependent.
Q5: $\Delta H = \text{negative}, \Delta S = \text{negative}$. Options: temp dependent.
Q6: $\Delta H = \text{positive}, \Delta S = \text{positive}$. Options: temp dependent.

Okay, so Q3 is the odd one out because it gives $\Delta G$ directly.
If $\Delta G$ is positive, it is non-spontaneous.
If $\Delta S$ is negative, and we assume standard conditions where this pair ($\Delta H>0, \Delta S<0$) is discussed, it is non-spontaneous at all temps.
Answer for 3: b. always non-spontaneous at all temperatures (Assuming the question implies determining the nature of the reaction based on the signs).

4. Given the following equation circle the correct answer:
$\Delta G = \text{negative}$
$\Delta S = \text{positive}$

* What is $\Delta H$?
Again, the options describe spontaneity.
If $\Delta G$ is negative, the reaction is spontaneous.
If $\Delta S$ is positive, the entropy change favors spontaneity.
Typically, this pairs with $\Delta H$ being negative (exothermic). If $\Delta H < 0$ and $\Delta S > 0$, $\Delta G$ is negative at all temperatures.
Answer: a. always spontaneous at all temperatures

5. Given the following equation circle the correct answer:
$\Delta H = \text{negative}$
$\Delta S = \text{negative}$

* What is $\Delta G$?
Formula: $\Delta G = \Delta H - T\Delta S$
$\Delta H$ is negative (favors spontaneity).
$\Delta S$ is negative (opposes spontaneity, makes $-T\Delta S$ positive).
At low temperatures, the $\Delta H$ term dominates, so $\Delta G$ is negative (spontaneous).
At high temperatures, the $-T\Delta S$ term becomes very large and positive, overcoming $\Delta H$, so $\Delta G$ becomes positive (non-spontaneous).
Answer: c. negative at lower temperatures, positive at higher temperatures (which means spontaneous at low temps, non-spontaneous at high temps).

6. Given the following equation circle the correct answer:
$\Delta H = \text{positive}$
$\Delta S = \text{positive}$

* What is $\Delta G$?
Formula: $\Delta G = \Delta H - T\Delta S$
$\Delta H$ is positive (opposes spontaneity).
$\Delta S$ is positive (favors spontaneity, makes $-T\Delta S$ negative).
At low temperatures, the positive $\Delta H$ dominates, so $\Delta G$ is positive (non-spontaneous).
At high temperatures, the $-T\Delta S$ term becomes very large and negative, overcoming $\Delta H$, so $\Delta G$ becomes negative (spontaneous).
Answer: d. negative at higher temperatures, positive at lower temperatures (which means spontaneous at high temps, non-spontaneous at low temps).

7. What must be true about the free energy of a reaction if the reaction is spontaneous?
The Gibbs Free Energy change ($\Delta G$) must be negative ($\Delta G < 0$).

8. What must be true about the free energy of a reaction if the reaction is NOT spontaneous?
The Gibbs Free Energy change ($\Delta G$) must be positive ($\Delta G > 0$).

*(Note: If $\Delta G = 0$, the reaction is at equilibrium).*

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Final Answer:

1. Spontaneous means a reaction happens naturally on its own without needing continuous outside energy.
2. Non-spontaneous means a reaction will not happen on its own and requires outside energy or work to occur.
3. b. always non-spontaneous at all temperatures
4. a. always spontaneous at all temperatures
5. c. negative at lower temperatures, positive at higher temperatures
6. d. negative at higher temperatures, positive at lower temperatures
7. $\Delta G$ must be negative ($\Delta G < 0$).
8. $\Delta G$ must be positive ($\Delta G > 0$).