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Problem 53

Find $\Delta G^{\circ}$ for the reactions in Prob…

Problem 52

Calculate $\Delta G^{\circ}$ for each reaction using $\Delta G_{\mathrm{f}}^{\circ}$ values:
(a) $\mathrm{H}_{2}(g)+\mathrm{I}_{2}(s) \longrightarrow 2 \mathrm{HI}(g)$
(b) $\mathrm{MnO}_{2}(s)+2 \mathrm{CO}(g) \longrightarrow \operatorname{Mn}(s)+2 \mathrm{CO}_{2}(g)$
(c) $\mathrm{NH}_{4} \mathrm{Cl}(s) \longrightarrow \mathrm{NH}_{3}(g)+\mathrm{HCl}(g)$

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Video Transcript

in each part of this problem were given a chemical reaction. And we need to calculate the change in Gibbs free energy of the reaction at standard conditions. This is the formula that we're going to apply to this problem. We first need to find the total change, Gibbs free energy of the products of the reaction. Then subtract the total change in Gibbs free energy of the reactant. So starting with part A, we're given the formation reaction of H two plus I to to form to H I. We can break up the terms in the formula, do you to be the products and the reactant since he and are we start by looking at the product side of the given formation reaction. We see that we have two moles of h I in the molds destroy geometrically efficient corresponds to this end in this formula. And so that's why we have this to here. And we want to play that by the Change and Gibbs free energy that standard conditions of formation for that at compound and we can look that up in the appendix we see that comes out to this value and when we multiply these two Together we're left with units of killer jewels or total energy. Since this is a formation reaction, the the elements of hydrogen and iodine form H I and they exist naturally as h two and I to. Because of that, they do not require any energy to change to form since they occur naturally. So that means that the change in Gibbs free energy of formation at standard conditions for each one of those those die Atomics will be zero. And when we add that together we get the total Gibbs free energy change of the reactant to be zero. So when we do, P minus are are a zero. So the total change in Gibbs free energy of their reaction is just equal to the total change and energy of the products. And again, that is 2.6 Killer Jewell's. So that's how we we work through these and we use this formula now to go through Parts B and C. This is the given reaction in part B and again we start with the products and we see that we have one mole of solid manganese. However, as in part A, this is naturally forming and therefore has a changing Gibbs free energy information value at standard conditions of zero. And so we do not include it in in our formula and also on the products we see that we have two moles of co two. So we want to play those two moles by the changing Gibbs Free Energy a formation at Senior conditions or CO two to get the total change in Gibbs free energy of the products. And now we do the reactions. You have one more UMNO to in two moles of CEO and one mole of M N 0 to 2 moles of CEO, each multiplied by their Delta G formation values gives us tee total changing Gibbs free energy of the reactors. And now we do the products minus the reactant. That is this. See total change in Gibbs Free Energy this reaction system which comes out to, you know, get a 48.3 killer JAL's and in part C. We have this given reaction. We're on the product side. We have one more of any age three and one mole of HCL after we multiply each by their respective delta G of formation values that standard conditions and add them together. We get this this value for the total energy change in the products. On the reactant side, we see that we just have a single mole of NH four c l. When we multiply it by its energy change value, we get the total. It's free energy change of of the reactant. Then again, we just do the products minus the reactant. And after that subtraction we get a delta G value of 91.7 killing jewels. The reaction in part C.