Gibbs Free Energy - Online Tutor, Practice Problems & Exam Prep

• Delta G measures energy change in a chemical/physical process
• Sign of delta G or K determines spontaneity of a reaction
• Gibbs free energy = 0 at equilibrium
• Delta G > 0 = non-spontaneous and K < 1 (reactants dominate)
• Delta G < 0 = spontaneous and K > 1 (products dominate)
• Delta H and delta S signs determine spontaneity when delta G unknown
• Delta H and delta S both positive = spontaneous at high temp
• Delta H and delta S both negative = spontaneous at low temp
• Delta H + and delta S - = always non-spontaneous
• Delta H - and delta S + = spontaneous

Spontaneity of Reactions

Now Gibbs free energy can be represented by ΔG. 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 ΔG and or the value of your equilibrium constant determine the spontaneity of a reaction. So if I knew the sign of ΔG, 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 0. Remember that K represents products over reactants, and we said that if products over reactants is equal to 1, this can signify being at equilibrium.

Now if Gibbs free energy happens to be a positive value, meaning it's greater than 0, 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 0, 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 their signs can help us determine if we are at equilibrium, spontaneous, or non-spontaneous.

Gibbs Free Energy Example

If ΔG 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 forward direction or the reverse direction. Now they're telling us ΔG is small, so that means it's close to 0 and it's positive, so that means it's slightly greater than 0, which means it's non spontaneous. And when we say that, we mean non spontaneous in the forward direction.

But remember we're dealing with reversible reactions when we talk about Gibbs 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&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 ΔG? When it's equal to zero we are at equilibrium. Since ΔG is small, that means it's a number that's pretty close to 0. So that means we would be near equilibrium. This would mean that option D is the correct answer.

Spontaneity and Temperature Conditions

When the sine of ΔG is unknown, we can say that the spontaneity of a reaction can predict it from the signs of enthalpy, which is ΔH, and entropy, which is ΔS. 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'd be spontaneous at low temperatures.

Next we're going to say when ΔH is positive and ΔS is negative, you are always non spontaneous in terms of your chemical reaction. And if we do the opposite of that, if ΔH is now negative and ΔS is positive then you'd be spontaneous for your chemical reaction.

So just remember, easiest way to remember this here your ΔH positive and negative. Huge ΔS positive and negative. So you'd say high T, low T, non and spawn. So the sides of our ΔH and ΔS for chemical reaction can be used to determine the right temperature conditions to make it spontaneous or not.

Gibbs Free Energy Example

Here we're going to say that CL₃ phosphorus trichloride plus chlorine gas react together to give us phosphorus pentachloride at 25°C. The enthalpy is -92.50 kilojoules. Which of the following statements is R true? This is an endothermic reaction. Well, ΔH here is negative, so it is exothermic, not endothermic.

If the temperatures increase, the ratio of our products over reactants will increase. So first of all, we're going to say products overreacted since 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 ΔH and ΔS. So ΔH positive, negative, ΔS positive and negative. We know that ΔH is negative because we're told that right from the beginning. So we're dealing with this one.

And then what you're going to say you can determine the sign of ΔS. We have two reactants combined to give us one product. So we're forming bonds, which means that ΔS is negative. So that means we'd fit in this slot, which means that our 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 going to have the opposite effect. It's going to 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, overreactants. 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 K, K would be decreasing. So this is not true here.

The ΔS for the reaction is negative, yes. What multiple reactants combine to give me one product that is a decrease in entropy. So it's negative. And ΔG 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.

The chemical reaction 2 NO₂Br (g) → 2 NO₂ (g) + Br₂ (g) has a K_eq = 4.50 × 10⁵.

Does the reaction increase the entropy of the Universe? Explain.

• A. Yes, because K_eq > 1 and ΔG < 0.
• B. No, because K_eq > 1 and ΔG > 0.
• C. Yes, because K_eq > 1 and ΔG = 0.
• D. No, because K_eq > 1 and ΔG = 0.

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 = (−)

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 = (−)

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.