ΔG˚=141.7 kJ for the following reaction. Calculate ΔG: T=10˚C, [SO 3 ] = 25mM, [SO 2 ] = 50mM, & [O 2 ] = 75 mM. 2 SO 3 (g) ⇌ 2 SO 2 (g) + O 2 (g)

Alright. So this practice problem wants us to calculate the actual change in Gibbs free energy, and it tells us that the standard change in Gibbs free energy is equal to a 141.7 kilojoules for the following reaction down below, where we have 2 moles of sulfur trioxide gas reacting to form 2 moles of sulfur dioxide gas and 1 mole of oxygen gas. And then it tells us that the temperature of the system is equal to 10 degrees Celsius, and it gives us the concentrations of all of the reaction components right here in units of millimolar. And so in order to calculate the actual change in Gibbs free energy, we're going to need to recall the equation that relates the actual change in Gibbs free energy to the standard change in Gibbs free energy. So I've gone ahead and provided that equation right here. And so, essentially, all we need to do to calculate the actual change in Gibbs free energy is to plug in all of these values right here into our calculators. And so, we know from our previous lesson videos that this R value right here is literally just the gas constant, which has a value of 8.315 joules per mole times Kelvin. And so all we need to do is plug in these other values. So we know that we're given the change in standard, in the standard change in Gibbs free energy, which is given to us as 141.7 kilojoules. So we can just take this number right here, a 141.7, and plug it in right here. Right? No. Not right. So, recall that we want the units in our expression to correspond and correlate with one another. And so, notice that the gas constant has units of joules. However, the standard change in Gibbs free energy is given to us in units of kilojoules, which means that we want to convert these units of kilojoules into units of joules, so that they match the, units in our gas constant. And so, if all we need to do to convert a 141.7 kilojoules into joules is to multiply this number by 1,000. And so that gives us an answer of 100 and 41,700 joules. So now we can take that, this value right here and plug it in to our standard change in Gibbs free energy. So we'll go ahead and plug that in here, a 141,700 and this is units of joules. So then, we can go ahead and plug in our temperature, and we're given the temperature of 10 degrees Celsius. So we can take this 10 degrees Celsius and plug it in right into our equation, right? No, not right. So remember that the temperature in this expression is in units of Kelvin, but we're given the temperature in units of degrees Celsius. So first, we need to take this temperature here and convert it into units of, Kelvin. So recall that Kelvin is equal to the degrees Celsius plus 273.15. And so, if we plug in the temperature of 10 degrees Celsius into this equation, we get that the Kelvin is equal to 283.15. And so now that we have our temperature in Kelvin, we can take it and plug it into our temperature in our expression. And so let's go ahead and do that. So, our temperature is going to be 283 0.15 Kelvin. And then, all of this is going to be, have this natural log of the reaction quotient at the very end. And so really, it's this reaction quotient that we don't have readily to plug in. So, we can actually calculate for this reaction quotient. So recall that the reaction quotient is literally just the ratio of the concentration of products over the concentration of reactants. And so, notice that we have 2 products here in our reaction. We have sulfur dioxide gas, so s o 2. So we can plug that in here, s o 2. And then we also have the oxygen gas here that we can plug in as a product. And then we also have the reactant, one reactant, which is the sulfur trioxide gas that can go down below. So, so far, this is looking good. Right? We're not missing anything very important. Right? No, not right. We're missing something incredibly important. Remember, that the coefficients that are in front of all of these, reactants need to be expressed as exponents. And so what that means is that notice that our sulfur dioxide right here has a coefficient of 2, so it needs to be expressed as an exponent up here of 2. The oxygen gas here, does not have a coefficient. The coefficient is 1, so we can just leave it blank here. And then the sulfur trioxide gas has a coefficient of 2, so we can plug that in as an exponent up here as 2. So now pretty much all we need to do is plug in the concentrations of each of these guys here into our expression, right? Well, it turns out you're right, we do need to plug in these, values here, but we need to also recognize that the concentrations are given to us in units of millimolar, and these units of millimolar do not correlate with the units that are in our gas constant. Notice that our gas constant doesn't have millimoles in it, instead it has moles. And so, essentially, what we wanna do is convert all of our concentrations here into units of molarity before we actually plug it into our expression. And so, in order to convert these values here of And so in order to convert these values here of millimolar into molarity, all we need to do is divide them by 1,000. And so if we start with our first product here, which is the sulfur dioxide gas, it's in 50 millimolar. So if we take 50 and divide it by a 1,000, we can convert it into molarity. So 50 divided by 1,000 comes out to 0.05, and this is molar. And don't forget to include the exponents that we had before, so it's squared. The oxygen gas has 75 millimolar, so if we divide that by a1000, we can turn it into 0 point 075 Molar. And then, of course, the sulfur trioxide, the sulfur trioxide, gas, reactant is 25 millimolar, but if we divide that by a1000, it comes out to 0.025 Molar, and of course, this is going to be squared from before. And so now, all we need to do is crunch these numbers in our calculator. So if you take 0 point 0.05 molar squared times 0.075 molar and divide it by 0.025 molar squared, you'll get an answer of 0 0.3 molar. And so recall again that molar is, units of moles per liter, but we're not really given a volume here, so we can assume that the volume is 1 liter, and then that allows us to assume that this, reaction quotient can be expressed in units of, moles here. So we have 0.3 moles. So now, we have our reaction quotient here that we can plug right into our expression. And so we can go ahead and plug in 0.3 into this expression, and now we have everything that we need. So, if you take your calculator and you do the natural log of 0.3 times 283.15 times 8.315, you'll get an answer of negative, so your answer will come out to negative 2834 point 62, and this will be in units of joules. And so we still have this value right here, a 141,700 joules, and this is being subtracted now. And so, essentially what we're saying is that all of this here, converts to this number here. And so all we need to do is combine these two numbers. So if you do a 141,700 joules minus 2834.62 joules, you'll get an answer of 100 138,865 joules about. And so this answer here matches with answer option d, so we can go ahead and indicate here that d is the correct answer for this problem. And so, because actual change in Gibbs free energy is a positive value here, it comes out to a positive value, that means that this is going to be an endergonic reaction. If it were to have come out as a negative value, then it would have been an exergonic reaction. And so that concludes this practice problem right here. And again, there's a lot of unit conversions, there's a lot of, places where you can make mistakes, but you wanna make sure that the units are all matching the units that are present in our gas constant. And so that, entails in our gas constant. And so that, entails converting the, delta g into joules, the concentrations into moles, and the temperature into Kelvin. And so, again, that concludes this practice problem, and I'll see you guys in our next video.

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1. Introduction to Biochemistry

Gibbs Free Energy

Biochemistry

1. Introduction to Biochemistry

Gibbs Free Energy

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Multiple Choice

ΔG˚=141.7 kJ for the following reaction. Calculate ΔG: T=10˚C, [SO ₃] = 25mM, [SO₂] = 50mM, & [O₂] = 75 mM.
2 SO₃(g) ⇌ 2 SO₂(g) + O₂(g)

– 6,437 J

39,938 J

-155,127 J

138,865 J

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Understand the relationship between standard Gibbs free energy change (ΔG˚) and actual Gibbs free energy change (ΔG) using the equation: ΔG = ΔG˚ + RT ln(Q), where R is the universal gas constant, T is the temperature in Kelvin, and Q is the reaction quotient.

Convert the temperature from Celsius to Kelvin by adding 273.15 to the given temperature: T(K) = 10 + 273.15.

Calculate the reaction quotient (Q) using the concentrations of the reactants and products: Q = ([SO2]^2 * [O2]) / ([SO3]^2). Substitute the given concentrations into this expression.

Use the universal gas constant R = 8.314 J/(mol·K) and the calculated temperature in Kelvin to compute the term RT ln(Q).

Substitute ΔG˚, RT ln(Q), and the calculated Q into the equation ΔG = ΔG˚ + RT ln(Q) to find the actual Gibbs free energy change (ΔG).

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Struggling with Biochemistry?

ΔG˚=141.7 kJ for the following reaction. Calculate ΔG: T=10˚C, [SO 3] = 25mM, [SO2] = 50mM, & [O2] = 75 mM. 2 SO3(g) ⇌ 2 SO2(g) + O2(g)

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Gibbs Free Energy practice set

ΔG˚=141.7 kJ for the following reaction. Calculate ΔG: T=10˚C, [SO ₃] = 25mM, [SO₂] = 50mM, & [O₂] = 75 mM.
2 SO₃(g) ⇌ 2 SO₂(g) + O₂(g)