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One of the steps in the extraction of iron from its ore (FeO) is the reduction of iron(II) oxide by carbon monoxide at $900^{\circ} \mathrm{C}:$ $\mathrm{FeO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Fe}(s)+\mathrm{CO}_{2}(g)$If $\mathrm{CO}$ is allowed to react with an excess of $\mathrm{FeO}$, calculate the mole fractions of $\mathrm{CO}$ and $\mathrm{CO}_{2}$ at equilibrium. State any assumptions.

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Chemistry 102

Chapter 17

Entropy, Free Energy, and Equilibrium

Thermodynamics

University of Central Florida

University of Kentucky

Brown University

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.

06:58

In the recovery of iron fr…

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Iron ore, Fe2O3(s), is tre…

02:58

The following reaction occ…

01:55

Consider the extraction of…

03:00

Iron oxide ores are reduce…

03:21

The balanced equation for …

02:23

Iron(III) oxide reacts wit…

09:40

03:02

In making iron from iron o…

02:34

So we have Ah, this reaction of iron oxide with carbon monoxide. And, uh, we want to find out the equilibrium constant s so that we can get the amounts of the carbon monoxide and the carbon dioxide and equilibrium. So we go to the appendix three and start looking up the free energies of formation. We take the sum for the products, the iron and the co two, and we subtract from it to some for the reactant Cynthy iron oxide in the carbon monoxide. And so I obtained each value zero for iron 3 94 negative. 3 94.4 for carbon dioxide and so on. Iron oxide and carbon monoxide. And when I put it all together, when I total up without making sign airs, I get a negative 1.9 chili jewels promote pretty close to zero. So we want to get the equilibrium constant. Uh, and so the equation that relates the equilibrium constant to the standard free energy change and dividing both sides by the negative rt gives me the log of K. And so I plug in the free energy change that I calculated standard for energy change and the value of our and this is said to be at 900 psi. So that's 11. 73 Kelvin's. I forgot to put my key there. I'm sorry. Ah, and I get a log of K that is equal to about 0.2. And so that means K is the exponential of that, giving me 1.21 And that is the value of the equilibrium constant, which is the ratio of the now that's a solid left out solid left out. Those don't go into the equilibrium, constant expression. It's the pressure of the co two over the pressure, their CEO. And so we'll let x be the more fraction of co two one minus X, the mole fraction of thes CEO and that is equal to 1.21 So multiplying both sides by when minus X, I get 1.21 times one and 1.21 times minus X is equal to X and x. The mole fraction of the CO two is 55% 4.55 and the more fraction of the CEO is 0.45

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