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The $K_{\mathrm{sp}}$ of $\mathrm{AgCl}$ is given in Table $16.2 .$ What is its value at $60^{\circ} \mathrm{C} ?[$ Hint: You need the result of Problem 17.51(a) and the data in Appendix 3 to calculate $\left.\Delta H^{\circ} \cdot\right]$

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$$2.6 \times 10^{-9}$$

Chemistry 102

Chapter 17

Entropy, Free Energy, and Equilibrium

Thermodynamics

Carleton College

Drexel 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.

01:49

Use Appendix D to calculat…

06:14

01:31

Use the thermodynamic data…

02:07

What is the $\Delta H_{\te…

02:22

Use the data in Appendix 3…

so I definitely took the hint. Uh, number one used appendix three to, um, calculate Thea Delta h for the reaction. And so that's the standard heat of formation of silver ion, plus the standard eight of formation of chloride ion minus the standard heat of formation of the silver chloride. And so I looked up the values from the table and I was careful about my signs, and I come up with 65 0.7 killer jewels promote. And then I wrote down the result, uh, in question 51 and I see that I have all the values for two temperatures and the delta H in the R and K 10 I need a skate to which is what was asked. So ah, I plugged in the delta H that I calculated here and are. And then I plugged in the two temperatures, which, as always, if you're actually using temperature, the values have to be in Kelvin our Kelvin's, I should say. And, uh, my result is that the log of the ratio of the case is 2.7 87 and therefore the ratio of the Cazes E to the 2.7 87. I was calculating the log by plugging in all of these values. Here. Here. Ah, and so basically, what I find is the K at the higher temperature will be 16 times the value that I looked up in the table. Uh, namely 2.6 times 10 to the minus nine.

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