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TBR Thermochemistry Confusion
- Thread starter Dochopeful13
- Start date
delta G = delta H - T*delta S
This is a fundamental equation you should know. Let's go through all the possibilities for signs of delta H and delta S (+/+, +/-, -/+, -/-). If both delta H and delta S are positive, delta G will get more negative with increasing temperature. This is because you're subtracting a larger positive number (T*delta S) from another positive number. If delta H is negative and delta S is positive, the same effect is observed. If both delta H and delta S are negative, delta G will get more positive with increasing temperature because you're subtracting a larger negative number (thus adding a large positive number) from a negative number. If delta H is positive and delta S is negative, the same effect is observed. Here, we see that delta G was favorable (overall negative) at 298 K but unfavorable (overall positive) at 473 K. Therefore, delta G got more positive and so the signs for delta H and delta S must be either -/- or +/-.
Now we know how delta G varies with temperature. But that doesn't tell us the sign of delta G - we only know whether it increases or decreases with temperature. The sign of delta G depends on the signs of delta H and delta S. So we have two possibilities here: -/- or +/-. So let's treat the second case first. Positive delta H and negative delta S. No matter what temperature you choose, delta G will be positive because T must be positive and delta S is negative, meaning that you add T*delta S = positive number to delta H, which is also positive. It's impossible to add a positive number to another positive number and get a negative number. So that means -/- is the correct answer. To see the derivation, let's go back to the fundamentals:
delta H - T*delta S < 0 (for spontaneous case)
delta H < T*delta S
delta H < 298 K*delta S
delta H < 298 K*(some negative number)
Since delta S is a negative number, delta H is less than some negative number, which must be a negative number itself.
This is a fundamental equation you should know. Let's go through all the possibilities for signs of delta H and delta S (+/+, +/-, -/+, -/-). If both delta H and delta S are positive, delta G will get more negative with increasing temperature. This is because you're subtracting a larger positive number (T*delta S) from another positive number. If delta H is negative and delta S is positive, the same effect is observed. If both delta H and delta S are negative, delta G will get more positive with increasing temperature because you're subtracting a larger negative number (thus adding a large positive number) from a negative number. If delta H is positive and delta S is negative, the same effect is observed. Here, we see that delta G was favorable (overall negative) at 298 K but unfavorable (overall positive) at 473 K. Therefore, delta G got more positive and so the signs for delta H and delta S must be either -/- or +/-.
Now we know how delta G varies with temperature. But that doesn't tell us the sign of delta G - we only know whether it increases or decreases with temperature. The sign of delta G depends on the signs of delta H and delta S. So we have two possibilities here: -/- or +/-. So let's treat the second case first. Positive delta H and negative delta S. No matter what temperature you choose, delta G will be positive because T must be positive and delta S is negative, meaning that you add T*delta S = positive number to delta H, which is also positive. It's impossible to add a positive number to another positive number and get a negative number. So that means -/- is the correct answer. To see the derivation, let's go back to the fundamentals:
delta H - T*delta S < 0 (for spontaneous case)
delta H < T*delta S
delta H < 298 K*delta S
delta H < 298 K*(some negative number)
Since delta S is a negative number, delta H is less than some negative number, which must be a negative number itself.
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Yeah, for these questions the best thing to do is to think of the right side of the equation ∆G = ∆H - T∆S as having TWO components. The first component is ∆H. The effect of ∆H is obvious - a more negative ∆H is more favorable. The second component is (-T∆S). The negative sign makes things a bit screwy here, so think of it like this. This component is a fudge factor which will either help or oppose the effect of the ∆H. The EFFECT of the fudge factor (how much it helps or opposes ∆H) is dependent on two things - 1) Sign (determined by ∆S) 2) Magnitude (determined by T)
Sign: If ∆S is positive, the entire fudge factor is NEGATIVE. If ∆S is negative, the fudge factor is POSITIVE. The negative fudge factor (positive ∆S) will HELP a negative ∆H be more negative, it will make the ∆G more negative, it will make the reaction favorable (this makes sense, because more entropy is favorable). A positive fudge factor (negative delta H) will HELP a positive ∆H be more positive, and the reaction will be less favorable.
Magnitude: If T is large, the entire fudge factor will be large. Thus, the fudge factor will have a LARGE effect. Vice versa if it is small. So if T is large and ∆S is positive, the entire fudge factor will be a LARGE negative value. If ∆H is negative, the ∆G will then be very negative. If ∆H is positive however, the value of ∆H must be VERY high to overcome the effect of the LARGE negative fudge factor.
The thing that used to trip me about this was the question "what if ∆H is positive, T is small, but ∆S is VERY positive? Couldn't that be spontaneous" The answer is no. You can see this from looking at the units. ∆S is in J/mol, while ∆H (and ∆G) are in kJ/mol. Also important to remember is that "T" will never be negative.
So for ∆S, the sign matters but the magnitude of the entropy change really doesn't. What we care about is the sign of the two components of the system. If both components are negative (∆H and -T∆S), the reaction is spontaneous at all temperatures (Again, -T∆S is negative when ∆S is positive). If both are positive (+∆H & -∆S), the reaction is non-spontaneous at all temperatures. In both these cases, the ∆H and the -T∆S components create the same effect.
If, however, the two components oppose each other, like when both ∆H and ∆S are negative, then the fudge factor (-T∆S) will be positive. If the temperature is low, the effect of the fudge factor will be low, so ∆G will be negative at low temperatures. Raising the temperature however will increase the effect of the negative entropy change, and once the magnitude of T∆S is greater than ∆H, the reaction will no longer be spontaneous.
If both ∆H and ∆S are positive, then the fudge factor will be negative (favorable). This means that at low temperatures, the effect of the fudge factor will be small, so the positive ∆H will win the battle and the reaction will be non-spontaneous. Increasing the temperature will increase the effect of the -T∆S value, and when the magnitude exceeds the magnitude of ∆H, then the reaction will be spontaneous (∆G < 0).
Sign: If ∆S is positive, the entire fudge factor is NEGATIVE. If ∆S is negative, the fudge factor is POSITIVE. The negative fudge factor (positive ∆S) will HELP a negative ∆H be more negative, it will make the ∆G more negative, it will make the reaction favorable (this makes sense, because more entropy is favorable). A positive fudge factor (negative delta H) will HELP a positive ∆H be more positive, and the reaction will be less favorable.
Magnitude: If T is large, the entire fudge factor will be large. Thus, the fudge factor will have a LARGE effect. Vice versa if it is small. So if T is large and ∆S is positive, the entire fudge factor will be a LARGE negative value. If ∆H is negative, the ∆G will then be very negative. If ∆H is positive however, the value of ∆H must be VERY high to overcome the effect of the LARGE negative fudge factor.
The thing that used to trip me about this was the question "what if ∆H is positive, T is small, but ∆S is VERY positive? Couldn't that be spontaneous" The answer is no. You can see this from looking at the units. ∆S is in J/mol, while ∆H (and ∆G) are in kJ/mol. Also important to remember is that "T" will never be negative.
So for ∆S, the sign matters but the magnitude of the entropy change really doesn't. What we care about is the sign of the two components of the system. If both components are negative (∆H and -T∆S), the reaction is spontaneous at all temperatures (Again, -T∆S is negative when ∆S is positive). If both are positive (+∆H & -∆S), the reaction is non-spontaneous at all temperatures. In both these cases, the ∆H and the -T∆S components create the same effect.
If, however, the two components oppose each other, like when both ∆H and ∆S are negative, then the fudge factor (-T∆S) will be positive. If the temperature is low, the effect of the fudge factor will be low, so ∆G will be negative at low temperatures. Raising the temperature however will increase the effect of the negative entropy change, and once the magnitude of T∆S is greater than ∆H, the reaction will no longer be spontaneous.
If both ∆H and ∆S are positive, then the fudge factor will be negative (favorable). This means that at low temperatures, the effect of the fudge factor will be small, so the positive ∆H will win the battle and the reaction will be non-spontaneous. Increasing the temperature will increase the effect of the -T∆S value, and when the magnitude exceeds the magnitude of ∆H, then the reaction will be spontaneous (∆G < 0).
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The thing that used to trip me about this was the question "what if ∆H is positive, T is small, but ∆S is VERY negative? Couldn't that be spontaneous" The answer is no. You can see this from looking at the units. ∆S is in J/mol, while ∆H (and ∆G) are in kJ/mol. Also important to remember is that "T" will never be negative.
The answer is it depends on what "small" is for T and how large delta S is. Usually you get the endothermic but exergonic reactivity in cases where you have large increases in entropy - like when you turn things into gases. Two real-world examples are dry ice and cold packs. If you've ever held dry ice in your hand, you'll notice that your hand gets really cold. As the dry ice sublimes from solid to gas, it takes heat from your hand (endothermic). Yet, the process is spontaneous above -78 centigrade or so. Why? Because there is a large entropy increase going from solid directly to gas that more than makes up for the endothermicity of the transformation.
Cold packs are another story. Cold packs contain two pouches separated by a membrane. In one pouch is water and in the other is a chemical whose dissolution in water is endothermic (like ammonium nitrate). When you break the cold pack, the membrane separating these pouches is broken and the ammonium nitrate is solvated in water spontaneously. It does so spontaneously because even though delta H is positive, the resulting entropy increases a lot and therefore the cold pack feels cold to the touch - it's taking heat from the water and increasing the chemical disorder within, making the overall process exergonic.
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The answer is it depends on what "small" is for T and how large delta S is. Usually you get the endothermic but exergonic reactivity in cases where you have large increases in entropy - like when you turn things into gases. Two real-world examples are dry ice and cold packs. If you've ever held dry ice in your hand, you'll notice that your hand gets really cold. As the dry ice sublimes from solid to gas, it takes heat from your hand (endothermic). Yet, the process is spontaneous above -78 centigrade or so. Why? Because there is a large entropy increase going from solid directly to gas that more than makes up for the endothermicity of the transformation.
Cold packs are another story. Cold packs contain two pouches separated by a membrane. In one pouch is water and in the other is a chemical whose dissolution in water is endothermic (like ammonium nitrate). When you break the cold pack, the membrane separating these pouches is broken and the ammonium nitrate is solvated in water spontaneously. It does so spontaneously because even though delta H is positive, the resulting entropy increases a lot and therefore the cold pack feels cold to the touch - it's taking heat from the water and increasing the chemical disorder within, making the overall process exergonic.
Super interesting, thanks for the insight. Would you say that on the MCAT, in the context of a question like the one in the OP, it is safe to assume that the magnitude of ∆S will be insignificant?
Super interesting, thanks for the insight. Would you say that on the MCAT, in the context of a question like the one in the OP, it is safe to assume that the magnitude of ∆S will be insignificant?
I don't think that's a safe assumption at all. Always return to the fundamentals, the Gibbs equation in this case. Start your reasoning from there and you won't go wrong. You have the equation, which has four variables. If the question is asking you about one of the variables, then that means information to figure out what matters for the other three variables is either in the question stem or passage.
