Calculate ΔG° at 298 K for these reactions and predict the effect on ΔG° of lowering the temperature. a. NH 3 (g) + HBr(g) → NH 4 Br(s) b. CaCO 3 (s) → CaO(s) + CO 2 (g) c. CH 4 (g) + 3 Cl 2 (g) → CHCl 3 (g) + 3 HCl(g) (ΔG° f for CHCl 3 (g) is -70.4 kJ/mol.)

Welcome everybody. Our next problem says calculate delta G standard at 298 Kelvin for this reaction and predict the effect on delta G of lowering the temperature. The reaction is CC three solid becomes co solid plus co two gas. Our answer choices are delta G equals negative 101.4 kilojoules. Delta G will be the same B delta G equals positive 131.4 kilojoules delta G will become more positive choice. C delta G equals negative 131.4 kilojoules delta G will become more negative and choice D delta G equals positive 101.4 kilojoules delta G will become zero. So we're looking for two things in this problem, we need to calculate a delta G standard value for the reaction. So we need delta G standard for the reaction and we need to predict what effect lowering the temperature will have on delta G. So we have an equation for the relationship between delta G and the temperature, which is that delta G standard is equal to delta H standard minus T delta S standard. Now notice we don't actually have any entropy values here. Our delta H. But we don't need to use this whole, we don't need to actually fill out this equation. We just need to look at the relationship between temperature and delta G. So what happens in general, the general trend if I lower the temperature? So let's first start by solving for our delta G of reaction, which we can solve, we don't use our delta H minus T delta S formula since we don't have any uh H values. But we have another formula we can use for delta G, which is that delta G of the reaction equals the sum of N times delta G of formation of the products minus the sum of N times delta G of formation of the reactants. Those delta G of formation are values. We can just look up for each of our reactants and products. So we can get all of those values and recall the N is the number of moles based on the coefficients in our equation. So first, we also need to double check is our equation balanced. This is a pretty simple one. And sure enough, we can see we have one calcium on either side, three oxygens, on either side and one carbon on either side. So we are balanced, all our coefficients are one. So keeping that in mind, so let's solve for this. So our products, we have two of them. So our delta G of reaction will equal. So we have our first product cao so we have one mole. We don't need the one in the sense that multiplying by one isn't going to affect our number, but it will impact our units because as we can see when we look up the delta geo formation, let's look up these values. So for C AC 03, put a little list here, delta geo formation, we have negative 1129.1 kilojoules per mole. I'm gonna put parentheses around this value. I have a negative sign there. I don't want to lose track of the fact that I need that. Otherwise I'll really mess myself up if I missed that negative cao negative 603 0.3 kilojoules per mole CO2 negative 394.4 kilojoules per mole. And again, you can look these up in a table. So let's go ahead and plug them in. And now you can see why I need that one mole in there because I need to cancel out the per moles in my delta geo formation values. So again, I'm looking at products. So for cao calcium oxide, I have negative 603.3 kilo jules for more, I'm gonna put a bracket around this and then plus the heated formation for my calcium, my excuse me, my carbon dioxide that will also be one mole and the heated formation there is negative 394.4 kilojoules per more. And all of that will then be minus the sum of the N delta G formation of the reactants. We only have one reactant C AC 03. So we subtract one more multiplied by my heated formation or my, excuse me, my free energy formation, not heated formation, 1129.1 kilojoules per mole. So we'll go ahead and simplify this a little bit combining our values for our products. And we see the moles have canceled out in each of my terms here. And so simplified, I have negative 997.7 kilojoules less because I subtracted a negative number 1129.1 kilojoules. And that is going to equal overall positive 131 0.4 kilojoules and just sort of a gut check there. Does that make sense? Yes, because my positive number here is larger than my negative number. So we have our answer choice here or value here because each of my different values for delta G are different. Once I've arrived at this, I honestly if I were on a test and in a hurry, don't even need to go any farther. So I have delta G is positive 131.4 kilojoules. So I know that choice B which has that answer is my answer. But since this is a practice problem, we want to be thorough here, we'll work through the effect on delta g of lowering the temperature. So we return to our equation here of delta G standard equals delta H standard minus T delta S standard. So we need to think about this term negative T delta S. So what do we know about delta S the change in entropy for this reaction? Well, we know we're going from one molecule of reactant, 22 molecules of product. Anytime we increase the number of molecules, we increase the entropy of our reaction. So we know that delta S standard for this reaction must be positive since entropy is increasing. So if delta S is positive, then negative T delta S remains negative. So we are subtracting the T delta S term, which T delta S is positive. So we're subtracting a positive value from R delta H value. So if that's the case, what happens as our T value becomes smaller? So as T decreases, then T delta S becomes smaller. So we are subtracting a smaller positive amount from our delta H. So this means that our delta G will become more positive. And that might be easier to conceptualize if you think of the fact that you are adding a negative term, so the smaller the negative term you're adding the more positive your final answer. So again, we saw in the answer that we already picked that it says right there delta G will become more positive. So choice B is definitely our answer. And just a quick word about our other answer. Choices here. Obviously, in choice, a delta G will be the same. This would not be expected because delta G is dependent on temperature, it varies with temperature. So you wouldn't expect it to be the same since we know that changing the temperature changes the value. So that one you could eliminate right away. And on choice D delta G will become zero. Well, it can become zero but only under a very specific condition where delta H is equal to T delta S. So this is only at one specific temperature that this happens. We're asked in general as you lower T. So that would definitely not be our correct answer even if it were correct value of delta G. So just a quick word on those two there and then obviously choice C delta G will become more negative is the opposite effect. So choice C also incorrect again, although we were able to determine that just from the value of delta G in this case. So once again, for calculating delta G standard at 298 K, for this reaction, predicting the effect on delta G of lowering the temperature, our answer is choice B delta G is positive 131.4 kilojoules and delta G will become more positive as we lower the temperature. See you in the next video.

Ch.18 - Free Energy and Thermodynamics

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Ch.18 - Free Energy and Thermodynamics

Problem 94

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Chapter 18, Problem 94

Calculate ΔG° at 298 K for these reactions and predict the effect on ΔG° of lowering the temperature.
a. NH₃(g) + HBr(g) → NH₄Br(s)
b. CaCO₃(s) → CaO(s) + CO₂(g)
c. CH₄(g) + 3 Cl₂(g) → CHCl₃(g) + 3 HCl(g) (ΔG°_f for CHCl₃(g) is -70.4 kJ/mol.)

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Identify the reaction: CaCO_3(s) \rightarrow CaO(s) + CO_2(g).

Use the standard Gibbs free energy change formula: \( \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \).

Look up the standard enthalpy change (\( \Delta H^\circ \)) and standard entropy change (\( \Delta S^\circ \)) for each substance involved in the reaction from a data table.

Calculate \( \Delta H^\circ \) and \( \Delta S^\circ \) for the reaction using the formula: \( \Delta X^\circ = \sum \Delta X^\circ_{\text{products}} - \sum \Delta X^\circ_{\text{reactants}} \), where \( X \) is either enthalpy or entropy.

Substitute the values of \( \Delta H^\circ \), \( \Delta S^\circ \), and \( T = 298 \text{ K} \) into the Gibbs free energy formula to find \( \Delta G^\circ \). Consider the effect of lowering the temperature on \( \Delta G^\circ \) by analyzing the term \( -T\Delta S^\circ \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Gibbs Free Energy (ΔG)

Gibbs Free Energy (ΔG) is a thermodynamic potential that measures the maximum reversible work obtainable from a thermodynamic system at constant temperature and pressure. It is a crucial concept in predicting the spontaneity of a reaction; a negative ΔG indicates a spontaneous process, while a positive ΔG suggests non-spontaneity. The standard Gibbs free energy change (ΔG°) is calculated under standard conditions, providing a reference point for reactions.

Temperature's Effect on ΔG

The temperature of a system can significantly influence the Gibbs free energy change (ΔG) of a reaction. According to the Gibbs-Helmholtz equation, ΔG is dependent on both enthalpy (ΔH) and entropy (ΔS) changes of the reaction. Lowering the temperature generally decreases the contribution of the entropy term (TΔS) to ΔG, which can lead to an increase in ΔG for reactions that are entropy-driven, potentially shifting the reaction from spontaneous to non-spontaneous.

Reaction Quotient (Q) and Equilibrium

The reaction quotient (Q) is a measure of the relative amounts of products and reactants present in a reaction at any point in time, compared to their equilibrium concentrations. It helps in determining the direction in which a reaction will proceed to reach equilibrium. For the reaction CaCO3(s) → CaO(s) + CO2(g), understanding Q is essential for predicting how changes in temperature and concentration will affect the spontaneity and equilibrium position of the reaction.