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The following diagram shows the variation of the equilibrium constant with temperature for the reaction $\mathbf{I}_{2}(g) \rightleftharpoons 2 \mathbf{I}(g)$ Calculate $\Delta G^{\circ}, \Delta H^{\circ},$ and $\Delta S^{\circ}$ for the reaction at 872K. (Hint: See Problem 17.51.)

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$\Delta G^{\circ}=62.5 \mathrm{kJ} / \mathrm{mol} ; \Delta H^{\circ}=157.8 \mathrm{kJ} / \mathrm{mol} ; \Delta S^{\circ}=109$$\mathrm{J} / \mathrm{K} \cdot \mathrm{mol}$.

Chemistry 102

Chapter 17

Entropy, Free Energy, and Equilibrium

Thermodynamics

Carleton College

University of Maryland - University College

University of Kentucky

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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Consider the reaction:…

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Calculate the equilibrium …

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So we are referred to problems 17 51 where an equation is given that relates equilibrium constants at different temperatures to those different temperatures and to the standard entropy change eso We have the two equilibrium constants at the two temperatures. And of course, the value of our is known. So we can just plug everything into this equation. The ratio of the K's is to 66.7, and then we will take the log of that, which is 5.5 86. Multiply it by our and ah, calculated this temperature factor and divided by it. And we get that the ah standard entropy change is a positive 1 57.8 And then, uh, at 8 70 to the value of K is given. And so we can plug in R and T and K and calculate Delta G zero and we get 62.5. And now the three properties Delta G. Delta Delta s are related. And so we can solve this equation for Delta s first subtracting Delta H from both sides and then dividing by the negative of tea. And when we do that, we get the value for, uh, the standard entropy change living off zeros. Here we go.

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