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The equilibrium constant $\left(K_{P}\right)$ for the reaction $\mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g)$ is 4.40 at 2000 K. (a) Calculate $\Delta G^{\circ}$ for the reaction. (b) Calculate $\Delta G$ for the reaction when the partial pressures are $P_{\mathrm{H}_{2}}=0.25 \mathrm{atm}, P_{\mathrm{CO}_{2}}=0.78 \mathrm{atm}, P_{\mathrm{H}_{2} \mathrm{O}}$ $=0.66 \mathrm{atm},$ and $P_{\mathrm{CO}}=1.20 \mathrm{atm}.$

$$-133 \times 10^{3} \mathrm{J} / \mathrm{mos}$$

Chemistry 102

Chapter 17

Entropy, Free Energy, and Equilibrium

Thermodynamics

Carleton College

University of Central Florida

Rice University

University of Kentucky

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.

04:20

The equilibrium constant $…

02:43

01:27

The following equilibrium …

00:57

05:14

Consider the reaction:…

02:47

For the following reaction…

01:04

At $425^{\circ} \mathrm{C}…

05:20

So we have this reaction Hold the substances in the reaction are gases on so we can write the Q expression from the reaction with the pressures of the products and the numerator pressures of the reactant and the denominator. We'll need this queue later. So I'm just going ahead and calculating it now and plugging in the values that are given in part B. We see that the Q is a little bit less than K. So as we know when Hugh is less than K in order to get equilibrium, que has got an increase. And that means Thebe products have to increase. And so the reactions going to go to the right on have a negative Delta G. But I look how close K is 4.4 more. We're basically 4.1, so it should be pretty small. Ah soas faras The Delta G zero goes. We calculate that from the log of the equilibrium constant natural law times minus rt and so plugging in our values. Ah, for our, uh, anti was given in Kelvin's Ah, we get a negative 24 0.6 colleges promote And so the reaction surprising with E. K just that small. Ah, less than five. We still get 24 killer jewels released now. Delta G zero is one thing. Delta G is equal to Delta G zero plus rt lnk Noticed the rt Ln um that, by the way, is equal to, uh, minus rt Ellen K over Q on. So really Delta G just gives you a mathematical comparison of K and cute like I just did. And in a qualitative way, just looking at the numbers anyway, plugging in the value we just calculated for Delta G zero and again R and T as given. And now the cue that we have calculated. I'm going to take the log of Q and then multiply by rt and we get a value that's pretty close. We've got a little over 24,000 jewels here, a little less than a little more than 23 here. So, uh, a little more than 1300 1.33 Killer jewels negative. Again, We do expect a negative value because Q has gotta increase in order, get to K

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