Gibbs Free Energy - Video Tutorials & Practice Problems
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1
concept
Spontaneity of Reactions
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Now gives free energy can be represented by delta G. Here, it's a measure of the energy change of a chemical or physical process that can be used to do work. Here, we say that the sign of delta G and or the value of your equilibrium constant, determine the spontaneity of a reaction. So if I know the sign of delta G, I can tell if a reaction is spontaneous. If I know the value of K, then I also know if a reaction is spontaneous. Now, here we're looking at non spontaneous equilibrium and spontaneous conditions. If we are at equilibrium, Gibbs free energy is equal to zero, remember that K represents products over reactants. And we said that if products overreact and S equal to one, this can signify us being equilibrium. Now, if Gibbs free energy happens to be a positive value, meaning it's greater than zero, that means we are non spontaneous. But it also means that your equilibrium constant is less than one where reactants dominate over products. Now, looking on the other side, if Gibbs free energy happens to be negative less than zero, then we are spontaneous. When we're spontaneous, we can also say that our equilibrium constant is greater than one, there's more products than reactants. So keep this in mind knowing both of their signs or either of those signs can help us determine if we are at equilibrium spontaneous or non spontaneous.
2
example
Gibbs Free Energy Example
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If delta G is small and positive, which of the following statements is true. So before we solve this question, realize that we're dealing with reversible reactions so they can go in the four direction or the reverse direction. Now they're telling us delta G is small. So that means it's close to zero and it's positive. So that means it's slightly greater than zero, which means it's non spontaneous. And when we say that we mean non spontaneous in the forward direction. But remember we deal with reversible reactions when we talk about gifts free energy. So if you're non spontaneous in the forward direction, that means that you are spontaneous in the reverse. So let's see what makes the most sense. So it's positive. So that means it's going to be non spontaneous in the forward, which means that A and B are out because they wouldn't be spontaneous in the forward, it will be spontaneous though in the reverse, which is what we said. But now is it far from equilibrium or near equilibrium? Remember delta G, when it's equal to zero, we are at equilibrium since delta G is small, that means it's a number that's pretty close to zero. So that means we would be near equilibrium. This would mean that option D is the correct answer.
3
concept
Spontaneity and Temperature Conditions
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When the sine of delta G is unknown, we can say that the spontaneity of a reaction can predict it from the signs of entropy, which is delta H and entropy which is delta S. So here, we're going to say that when they're both positive, we are spontaneous at high temperatures. And if both of them being positive makes it spontaneous at high temperatures, then what happens when it's the opposite? Well, when they're both negative, we be spontaneous at low temperatures. Next, we're going to say when delta H is positive and delta S is negative, you are always non spontaneous in terms of your chemical reaction. And if we did the opposite of that, if delta FH is now negative and delta S is positive, then you'd be spontaneous for your chemical reaction. So just remember the easiest way to remember this here, your delta H positive and negative, huge delta S positive and negative. So you'd say high t, low T non and spawn. So the size of our delta H and delta S for chemical reaction can be used to determine the right temperature conditions to make it spontaneous or not.
4
example
Gibbs Free Energy Example
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Here, we're going to say that PC L three phosphorus trichloride plus chlorine gas react together to give us uh phosphorus PTA at 25 °C. The entropy is negative 92.50 kilojoules. Which of the following statements is, are true. This is an endothermic reaction. Well, delta H here is negative. So it is exothermic, not endothermic. If the temperatures increase the ratio of our products over reactants will increase, right? So first of all, we're going to say products over reactants is talking about our equilibrium expression. So we're talking about our equilibrium constant K. Next thing we're going to say is let's determine the sign of delta H and delta S. So delta H positive, negative delta S positive and negative, we know that delta H is negative because we're told that right from the beginning. So we're dealing with this one and then what you're gonna say, you can determine the sign of delta S. We have two reactants combined to give us one product. So we're forming bonds, which means that delta S is negative. So that means we fit in this slot, which means that our temp as temperatures decrease we become more spontaneous. OK. So we're spontaneous at low temperatures. So if we're increasing the temperature, we're increasing the temperature, that's gonna have the opposite effect. It's gonna make us less spontaneous and more non spontaneous. And realize when we're making our reaction more non spontaneous, that means it doesn't want to happen. That means the reverse direction is favored. So that means that reactants are favored more than products. OK. So remember K equals products over reactants. We're heading in the reverse direction where reactants become more favored. So your bottom part is becoming bigger and as a consequence, your top part is becoming smaller. So overall what's happening to Kay K would be decreasing. So this is not true here. The delta S for the reaction is negative. Yes. What multiple reactants combined to give me one product that is a decrease in entropy. So it's negative and delta G for the reaction has to be negative at all temperatures. No, it's only going to be negative or spontaneous at low temperatures. So the only statement here that's true would be option C.
5
Problem
Problem
The chemical reaction 2 NO2Br (g) → 2 NO2 (g) + Br2 (g) has a Keq = 4.50 × 105.
Does the reaction increase the entropy of the Universe? Explain.
A
Yes, because Keq > 1 and ΔG < 0.
B
No, because Keq > 1 and ΔG > 0.
C
Yes, because Keq > 1 and ΔG = 0.
D
No, because Keq > 1 and ΔG = 0.
6
Problem
Problem
What are the signs of ∆H, ∆S and ∆G for the spontaneous conversion of a solid into gas?
A
∆H = (+), ∆S = (+), ∆G = (−)
B
∆H = (+), ∆S = (+), ∆G = (+)
C
∆H = (−), ∆S = (+), ∆G = (−)
D
∆H = (+), ∆S = (−), ∆G = (−)
7
Problem
Problem
Consider the combustion of butane gas and predict the signs of ΔS, ΔH and ∆G.
C4H10 (g) + 13/2 O2 (g) ⟶ 4 CO2 (g) + 5 H2O (g)
A
ΔS = (−), ΔH = (+), ∆G = (+)
B
ΔS = (+), ΔH = (−), ∆G = (+)
C
ΔS = (+), ΔH = (+), ∆G = (−)
D
ΔS = (+), ΔH = (−), ∆G = (−)
8
Problem
Problem
You calculate the value of ΔG for a chemical reaction and get a positive value. Which would be the most accurate way to interpret this result?
a) If a mixture of reactants and products is created and left to equilibrate, the equilibrium mixture will contain more reactant than product.
b) If a mixture of reactants and products is created, we cannot say anything about its composition at equilibrium but we can say it will reach equilibrium very rapidly.
c) The reaction will not occur under any circumstances.
d) If a mixture of reactants and products is created and left to equilibrate, the equilibrium mixture will contain more product than reactant.
A
If a mixture of reactants and products is created and left to equilibrate, the equilibrium mixture will contain more reactant than product.
B
If a mixture of reactants and products is created, we cannot say anything about its composition at equilibrium but we can say it will reach equilibrium very rapidly.
C
The reaction will not occur under any circumstances.
D
If a mixture of reactants and products is created and left to equilibrate, the equilibrium mixture will contain more product than reactant.
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