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Predict the signs of $\Delta H, \Delta S,$ and $\Delta G$ of the system for the following processes at 1 atm: (a) ammonia melts at $-60^{\circ} \mathrm{C},$ (b) ammonia melts at $-77.7^{\circ} \mathrm{C},$ (c) ammonia melts at $-100^{\circ} \mathrm{C}$. (The normal melting point of ammonia is $\left.-77.7^{\circ} \mathrm{C} .\right)$

(a) $\Delta G<0 ; \Delta S>0 ; \Delta H>0$(b) $\Delta G=0 . \Delta S>0 ; \Delta H>0$$(\mathrm{c}) \Delta \mathrm{G}>0 . \Delta \mathrm{S}>0 ; \Delta \mathrm{H}>0$

Chemistry 102

Chapter 17

Entropy, Free Energy, and Equilibrium

Thermodynamics

Carleton College

Rice University

University of Toronto

Lectures

00:42

In thermodynamics, the zeroth law of thermodynamics states that if two systems are in thermal equilibrium with a third system, they are also in thermal equilibrium with each other.

01:47

A spontaneous process is one in which the total entropy of the universe increases. In a spontaneous process, the system will move from an ordered state to a disordered state, such as from ice to water, or from a solid to a gas. The concept of spontaneity was introduced by Rudolf Clausius in 1850.

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Predict the sign of $\Delt…

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Find $\Delta S_{\mathrm{rx…

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(a) What are the signs of …

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So we're melting ammonia here, and, uh, we want to so we can about the sign of the and put p entropy in free energy changes. So first of all, uh, if we were just presented with the reaction melting is going from solid to liquid, we would say, Well, that is going to have a positive entropy change. The liquid has Maurin trapeze in the solid dust, and that's not dependent on temperature. Ah, and then think about the process of melting. Um, you know, what do you have to do to put an ice cube To melt it, you have to put in the heat. So the Delta H Year is also going to be positive. Now, think about the effects of ah, both the Delta s and the Delta H being positive wealth. The Delta H being positive has a positive effect on Delta G, which is to say it tends to meet Dr G. More positive now until two h is positive. It's being multiplied by a positive value of tea and a negative. So the whole thing has a negative contribution to Delta G. That is makes Delta G more negative intense to make Delta G Ah, spontaneous and so higher temperature. The higher T is we've got a plus here. Remember, Iron T is thin, the more negative this value is gonna be. And so it low t Delta G is going to be, ah, positive because the, uh, entropy change doesn't again multiplied by negative t doesn't give enough of a negative to overcome the positive Delta H at the melting point where the melting point is defined as theme. The temperature at which the solid and liquid are at equilibrium, said Delta G is equal to zero at that temperature and then at any higher temperature notice have converted the Celsius temperatures D. Kelvin because all temperatures in equations air are in Kelvin's, uh, although not necessarily temperature changes. Um, now, ah, the temperature is high enough that when we multiply it by the positive delta s and then change the sign that that is going to be a greater negative on the positive value of Delta H and Delta G is negative at all. Temperatures above the melting temperature

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