Use standard free energies of formation to calculate ΔG° at 25 °C for each reaction in Problem 61. How do the values of ΔG° calculated this way compare to those calculated from ΔH° and ΔS°? Which of the two methods could be used to determine how ΔG° changes with temperature?

Question

Accepted Answer

All right. Hi everyone. So this question says they use the standard free energies of formation to calculate Delta G standard at 25 °C. For the following reaction here we have one mole of sulfur reacting with one mole of 02 gas to produce. So two as a gas, how do the values of delta G standard calculated this way compared to those calculated from delta H standard and delta S standard, which of the two methods could be used to determine how delta G standard varies with temperature. So here are or here is a table with the relevant delta H standard, delta S standard and delta G standard values as well as four different answer choices labeled A through D. Alright. So first let's start by calculating delta G standard using the standard free energies of formation provided on the right side of the tape. In order to find the delt G standard of a given reaction recall that you have to first take the sum of the standard free energies of formation for all of the products and then subtract that by the sum of all delta delta G standard values or the reactants keep in mind also that each of the values for delta G standard must be multiplied by their stoichiometry coefficient. So let's go ahead and get started for delta C standard of the reaction. We happen to have only one product which is so two. So in our balanced chemical equation, we had one mole of so two with a delta G standard of negative 300.1 kilo Joel per month. And so we're going to subtract this by the sum of the Delta G standard values of the reactants multiplied by their coefficient. So for the reactants, we have sulfur and oxygen gas, sulfur has a coefficient of one in our balanced chemical equation. So we'll multiply that by delta G standard which is 236.7 kilojoules per mole or sulfur. Now, it just so happens that oxygen gas has a delta G standard of zero kilojoules per mole. So zero multiplied by a coefficient of one equals simply zero. So after evaluating this expression, it's worth mentioning that units of moles cancel out, which means that our answer for the delta G standard of the reaction is equal to negative 536.7 kg jules. So even though we have the value for delta G standard of the reaction at no point, did we ever use the temperature provided to us in the beginning of the video? So this method, even though it's going to give us the correct answer. This method does not actually account for any changes in temperature. Instead, our second method that involves delta H standard and delta S standard can be used to factor in the temperature because recall that delta G standard is also equal to delta H standard subtracted by T multiplied by delta as standard in which tea is the temperature in Kelvin. So using the other values provided in the table, let's go ahead and calculate delta H standard of the reaction. And delta s standard of the reaction. Now starting off with delta H standard of the reaction. The values given to us in our table are the values of delta H standard of formation for all of the relevant chemical compounds. So in this case, delta H standard of the reaction is the sum of the standard entropies of the products subtracted by the sum of the standard entropies of the reactants. So that's products subtracted by reactants once again multiplied by their sto geometric coefficients. So our only product here is so two which has a stroy metric coefficient of one more multiplied by its delta H standard which is negative 296.8 kg joules per mole. I'm going to briefly scroll up here just to make sure that that's accurate, which it is. So that concludes our products. So now let's go ahead and include our reactants for sulfur. We have one mole multiplied by its delta H standard. Which is if I scroll up here 277.2 kilo Joel per m. Now it just so happens that oxygen has a delta H standard equal to zero kilojoules per mole. So zero, multiplied by one mole equals zero. So after adding these quantities together or rather subtracting them, I should say delta H standard for the reaction is equal to negative 574 ketel joules. So now let's go ahead and find delta S standard for the reaction. In the case of delta S standard. Delta S standard for the reaction is equal to the sum of the product entropies subtracted by the sum of the reactant entropies. So for so two, we have one more multiplied by its delta S standard value, which happens to be 248.2 jules per mole. Kelvin. So we'll subtract this by the sum of the reactant entropies. For sulfur. We had one more multiplied by its entropy of 167.8 jewels per mole. Kelvin. So this is added to one mole of oxygen multiplied by its standard entropy of 205.2 jules per mole. Kelvin. Once again throughout these calculations, units of moles cancel out. So after combining all of these quantities together Delta S standard for the reaction is equal to negative 124.8 jewels per Kelvin since moles have canceled out. But for the sake of keeping our units consistent. Let's convert our units of jewels per Kelvin into kilojoules per Kelvin. Now, to do this recall that you must divide this quantity by 1000, which is equivalent to taking our decimal place and moving it three spaces to the left. So delta S standard for the reaction equals negative 0.1248 kilo jewels her Kelvin. So for a final step, let's go ahead and solve for delta G standard using the standard entropy and the standard entropy. So delta G standard is equal to delta H standard or negative 500 and 74 Quito Jules subtracted by the temperature which is 25 °C. But recall that temperature must be in units of Kelvin. So to convert this, I'm going to take my 25 °C and add 273.15. This gives us a temperature in Kelvin, specifically 298.15 Kelvin, which we then multiply by delta S standard or negative 0.1248 kilo jewels per Kelvin. And so our final answer using this method comes out to negative 536 0.79 kg Jules, which as it turns out is equal to the value of delta G standard calculated using the standard free energies of formation. So there you have it. If I go ahead and I scroll up here to consider our answer choices. The correct answer is going to be option B in the multiple choice because delta G standard is equal to negative 536.7 kg joules. Both methods produce the same result for Delta G standard. The method using Delta H standard and delta S standard could be used to determine how Delta G standard varies with temperature. And so with that being said, if you watch this video all the way through, thank you very much. And I hope you found this helpful.

Use standard free energies of formation to calculate ΔG° at 25 °C for each reaction in Problem 61. How do the values of ΔG° calculated this way compare to those calculated from ΔH° and ΔS°? Which of the two methods could be used to determine how ΔG° changes with temperature?

Verified Solution

Key Concepts

Standard Free Energy of Formation (ΔG°f)

Gibbs Free Energy Equation

Temperature Dependence of ΔG°