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Problem 54

Find $\Delta G^{\circ}$ for the reactions in Problem 20.52 using $\Delta H_{\mathrm{f}}^{\circ}$ and$S^{\circ}$ values.

Answer

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## Discussion

## Video Transcript

and each part of this problem. We are given a chemical reaction and we want to determine the change in Gibbs free energy of the given reaction. But we need to find it. Using using this method, we first need to find the belt h of reaction. So the change in entropy at standard conditions of the given reaction in the way that we find this is by first determining the total and pulpy change of the products and then subtracting the total entropy change of the reactant. We don't need to find the change in Tropea of the Reaction Delta s and we do that by finding the total entropy of the products and subtracting the total entropy of the reactant. After we have Delta H and Delta s values, we can use the equation at the bottom Delta G equals still teach minus T delta s and using a temperature of 298 kelvin which corresponds to standard conditions of 25 degrees Celsius, you find that delta G value. So that's a process that we're going to use to work through this problem where they were given a reaction and we first want to find Delta each of that reaction using the top equation, we need to multiply for the products and the reactant the number of moles of each species multiplied by the delta H of formation at standard condition values in order to get in order to get the total entropy change of the products and then subtract the total until be change of the reactant. So on the products of this given reaction, we see that we have two moles of h i and we can look up in the appendix the Delta H of formation value at standard conditions for h I. We multiply those two together we get the total entropy change of the products since that is the only product in this equation. And now for the little p change of the reactant both h two and I to exist naturally and do not release any energy in order to be formed and because of that, they're Delta H of formation. Values are both zero and so that means that the total entropy change of the reactant sin this reaction is zero. And so now that we found HP and HR, we subtract hp minus h r. But because h R. Zero, The Delta h of the overall reaction is just equal to the total entropy change of the products which this 51.8 killer JAL's. Now we use the middle equation to solve for the change in entropy of the given reaction. So, on the products again, we have two moles of h I. In this time in the appendix, we look up for h I the value of the standard Moeller entropy of of that compound. You want to play those values together? You can't sell off the units to get it in terms of jewels per kelvin or entropy units just as we cancelled off moles where the entropy to get energy units of killing jewels. So the total total entropy of the products in this reaction comes out to be this value and for the reactant. So we have one mole of H two on one mole of I to and these both have non zero entropy values as opposed to mental p values. And we can look those up in the appendix when we multiply them by one mole and add them together we get the total and true p change of the reactors and then we do SP the total entropy of the products minus s already total inter p of the reactive and this is S P and we subtract S r. We get this value for the the change in an entropy of the reaction and we need to divide this by 1000 so that we can have similar units of killer jewels. When we plug it all into this bottom equation, Delta G equals still to h minus t Delta s because Delta each ad units of killer jewels and Delta s head units of jewels per kelvin. So when we multiply by onto the negative 30 convert this into killing jewels, we plug it into that delta G equation units of Kelvin cancel out So that the Delta G of reaction as energy units of killing JAL's. And when we plugged that all in the change in Gibbs free energy of this reaction comes out to be 2.4 killer JAL's and this answer may be slightly different compared to the one calculated using the the change in Gibbs free energy of formation at standard condition values. In the other problem that examined these same three reactions, but it should be fairly close to the value that you got before. So now we just do that for the remaining two reactions in part B, we start by calculating the entropy change delta h of reaction earning. With the products you sue, we see that we have one mole of solid manganese but that is naturally occurring and has a delta h of formation value at standard conditions of zero. And so we ignore it. And the only other product is two moles of CO two. So we include that multiply it by its until P a formation value to get the total entropy change of the products and then for the reactant swe, have one mole of m n 02 Two moles of CEO might play each one of those by there, standard until pia formation, change values and add them together. Get the total entropy change of the reactions and we do the products minus the reactive entropy change Get the overall delta each of reaction in units of killing JAL's and now for the entropy Again, With the products we see, we have one mole of mn two moles of Co two one mole of MN in two moles of CO two an entropy values for the entropy value for mn is non zero, unlike the entropy value. So we include that in the equation for entropy and we can solve out for the total and true P of the products. And on the reactant again, we have one mole of M n 02 and two moles of CEO one mole of M N 0 to 2 moles of CEO what to play them by their respective standard entropy values and then added altogether get the total entropy of the reactant. Then we take the total entropy of the products and subtract the total entropy of the reactant. To get Delta s of reaction, which we again should divide by 1000. You get in units of killing jewels so that when we do, Delta G equals Delta H minus t Delta s again using 45.1 killer JAL's, which we calculated for Delta H using the temperature of 298 Kelvin it at standard conditions in the calculated value of Delta s. We get the final answer for Delta G to be about negative 48.4 killer jewels for part B imports. See, we examine our final reaction that were given. We start again by calculating Delta each of reaction where the entropy of the products we have One mole of NH three and one mole of HCL Multiply each by their respective delta h of formation at seeing her condition values and add them together get the total entropy change of the products and then we have one reactant one mole of NH four c l multiplied by its delta H of formation value of standard conditions in order to get the total entropy change of the reactors. Then we take the total entropy change of the products and subtract the total in Philippi change of the reactant. You get Delta H of reaction and units of killing JAL's And now for the entropy. Delta s of reaction. Begin look at the products and see we have one mole of NH 31 mole of hcl. One more with NH 31 mole of HCL each multiplied by their respective and true p standard entropy values And then we add them all together get the total and true p of the products in units of jewels per Kelvin And for the reactant. So we just have one mole of NH four c l multiplied by its standard Moeller entropy. We get the total and therapy of the reactant, and we take the total entropy of the products and subtracting total entropy of the reactivates. And we get Delta s of reaction. We divide by 1000 to convert it into units of killer jewels. To cancel off units in the Delta G equals Delta H minus t Delta s equation. And when we plug in the values that we just found for a delta, each and Delta s using a temperature of 298 killed in into that equation, we get the overall changing Gibbs free energy. The reaction at seeing her conditions for the reaction given in part c B 19

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