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Use the standard free energy data in Appendix G to determine the free energy change for each of the following reactions, which are run under standard state conditions and $25^{\circ} \mathrm{C}$. Identify each as either spontaneous or nonspontaneous at these conditions.(a) $\mathrm{C}(s, \text { graphite })+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)$(b) $\mathrm{O}_{2}(g)+\mathrm{N}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)$(c) $2 \mathrm{Cu}(s)+\mathrm{S}(g) \longrightarrow \mathrm{Cu}_{2} \mathrm{S}(s)$(d) $\mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)$(e) $\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g)$(f) $\mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{CaSO}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)$

$\Delta G=-394.36 \frac{k J}{m o l}$$\Delta G=175.2 \frac{k J}{m o l}$$\Delta G=-324.5 \frac{k J}{m o l}$$\Delta G=-57.1 \frac{k J}{m o l}$$\Delta G=-29.4 \frac{k J}{m o l}$$\Delta G=18.3$

Chemistry 102

Chapter 16

Thermodynamics

University of Central Florida

University of Maryland - University College

Brown University

Lectures

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A spontaneous process is o…

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Using data from Appendix $…

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Okay, We're going to calculate the free energy change for these reactions by looking at the values for the free energy change given in appendix C G and these I've written in, um, Green, Right under. And the units for all of these are gonna be killed, Jules Promo. So let's start solving them again. You want to remember that when solving these problems? One account for coefficients and to it, the formula essentially is the summation of the products minus estimation of the reactor. So it's hard off there. And you will. Cat Negative three niner 0.36 essentially minus zero. Which then will be a school we were answering. Negative three. Find 4.36 killed. Jules per roll. Moving on to part two again. You're just gonna follow the same rule. Don't forget the coefficient. So this is products minus reactant. So you're just gonna have to times 87.6. That's going to give you an answer of 1 75.2 Killings rules her role part. See? Same thing this time we have a product in the reactant that are non zero. So negative. 86.2 plus problem. Sorry, Let's go back here. And you a minus. So it's gonna be minus 238.25 equals negative. 324.5. Kill your sperm onto part, Dean. So now we have negative 897.5. Linus, the summation between Nick to 37.1 and negative 60 Corinne, Point. Sorry. That's gonna give us an answer of negative 57.1 village rules, per So you can see that all these problems follow the same pattern. So once you find these values in a panic so you can just go ahead and solve problems. This one will. You do three times to be 94.36 and truck got from the summation. Three times they give 1 37.15 plus 7 42.2 That's gonna leave us with negative 29 4 kill. It cools her full. And now the last part of the problem. Same thing. Don't forget your coefficients, so you're gonna have to times, um, they to 8.59 whole. Plus negative until it's gonna be a negative 1 to 2.0 and you're used to attract out from negative one. So I've been 97.45 and end up with danger. You teen 0.3 tillage Permal And now we've done the problem.

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