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### (a) Find $\Delta S_{\mathrm{rxn}}^{\circ}$ for th…

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Ohio State University

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Problem 40

(a) Find $\Delta S_{\mathrm{rxn}}^{\circ}$ for the formation of HI(g) from its elements.
(b) Calculate $\Delta S_{\text { univ }},$ and state whether the reaction is spontaneous
at 298 $\mathrm{K}$ .

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## Video Transcript

in part a of this problem. We want to determine the change in standard molar entropy or Delta s reaction of the formation reaction of each I gas from its naturally occurring elements. We know that h two gas and I two solid have a delta each of formation value of zero since they occur naturally. So therefore, those are the two elements that were going to be using to form H I and this formation reaction. So we begin by balancing out this formation reaction. We see that we need a coefficient of two h i to balance out the two iodine from I to two. Hydrogen is from H two and now we use this reaction to first find the total entropy of the products and then subtract the total entropy of the reactant in order to find the change in total standard entropy of the reaction. So we start with the products term and we see on the product side we form two moles of H i gas. So we have two moles of h I. We look up in the appendix what the standard more entropy value of h I is and then we cancel off units of moles. So we're left with total entropy units of jewels per Kelvin. Now we do the same for the reactant Sturm we see we have one mole of both H two and I to so one mole for each one of those. And we look up their standard more entropy values in the appendix and we multiply them by the number of moles and then add them together to get the total entropy of the reactant. And then we do e minus are where this is P and this is our and only subtract those two values. This is the answer that we get for Delta s of reaction. However, this is for the formation of two moles of h I. And for standard standard formation reactions, we are interested in the production of one mole of the desired compound so we can divide that answer by two. Since we have a striking metric coefficient of two moles of h I being produced and inviting that by two would would result in the change an entropy of the reaction for the production of one mole of H I. That comes out to about 83 0.0 Jules per killed him. And so that is he answered apart A An important be. We want to calculate Delta s of the universe as a result of this reaction and then determine whether or not this reaction will proceed spontaneously. So this is a reaction. This is the equation that we used to saw four Delta s of the universe. And we can look up values for the standard Moller change in Gibbs free energy for both the products in the reactant and using equation similar to the one that we used to find the standard more entropy values. And again, we're subtracting products minus reactant and that will give us this value for Delta G. And we're told that this is a 298 Kelvin. So we have everything that we need to solve four Delta s universe. We know that for formation reactions, the reactant swill have a change in Gibbs free energy of zero, so we can cancel off that term. So we only pay attention to the product side. We have two moles of each eye that were forming, and when we multiply that by its Gibbs free energy value, we see that we come up with 2.6 killer JAL's of energy for that toll change Delta G of reaction. We can rearrange this equation now to solve for still to us universe equals Delta G reaction that standard conditions invited by the negative temperature. And we just saw it for the Delta G of reaction you be you 0.6 killing jewels and we divide that by the temperature of negative 298 Kelvin. And that comes out to If we were to convert that unit of killing JAL's into jewels by multiplying it by 1000 it comes out to you. Negative. 8.7 Jules per Kelvin. But again, this is for the formation of two moles of H I. So we can divide that answer by two to get the formation of one mole of H I. So the Delta s universe for the formation of one mole of H I comes out to be about negative for 0.4 Jules per Kelvin. And now, in order to determine whether or not this reaction of spontaneous, we need to examine the sign of Delta as universe if Delta s universes positive than the reaction is spontaneous if it's negative than its non spontaneous. Conversely, if the sign of Delta G is is positive like it is here than its reaction is non spontaneous. And if it were negative and it would be spontaneous so we can say that this reaction is non spontaneous again because signed for Delta s universe is negative.