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Ch.15 - Chemical Equilibrium

Chapter 15, Problem 151

The F ¬ F bond in F2 is relatively weak because the lone pairs of electrons on one F atom repel the lone pairs on the other F atom; Kp = 7.83 at 1500 K for the reaction F21g2 ∆ 2 F1g2. (b) What fraction of the F2 molecules dissociate at 1500 K?

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Hello. In this problem we are told the equilibrium constant value for the foreign reaction is 0.686. When it is carried out at 1700 kelvin. The chlorine chlorine bond is relatively weak because the lone electron pairs on one chlorine atom repel the lone electron pairs on another chlorine atom. One. What fraction of the chlorine molecules break apart? 1700 kelvin equilibrium pressure of chlorine at 1700 kelvin is 0.350 atmospheres. So let's begin by writing the equilibrium constant expression. So this is in terms of partial pressures. So we have the partial pressure of atomic chlorine not squared all over the partial pressure of our reactant which is molecular chlorine. And we're told that this is equal to 0.686. And so let's then solve for the partial pressure of our atomic chlorine, it's equal to 0.686 times the partial pressure of molecular chlorine which we're told is 0.350 atmospheres. And so this works out to then 0.2401. So we can find in the partial pressure of chlorine by taking this charitable sides This works out then 0.49 atmospheres. So this is then the equilibrium partial pressure of atomic chlorine. Let's now make a ice table. So we have our reaction and then we have initial change in equilibrium. So we don't know the initial amount of molecular florian's. We'll call that a initially the atomic chlorine is zero. Our changes then minus X. And then plus two X, equilibrium is initial plus a change. So a minus X. And two X. So our equilibrium Partial pressure of atomic chlorine then is equal to 0.49 atmospheres. Which then based on our ice table, is equal to two x. This then tells us the X is equal to 0. atmospheres. You can then find the amount of molecular coin at the start, You know then that the partial pressure of molecular chlorine at equilibrium is equal to a -6, Which is equal to a -0. atmospheres. And we know that the partial pressure of chlorine and equilibrium is 0. atmospheres. And so solving for a then is equal to 0.595 atmospheres. And so we're trying to find the fraction of chlorine molecules break apart and so that will be equal to X. That amount that breaks part over the initial amount. So this works out then 0.245 Atmospheres divided by 0.595 atmospheres. And this first out to the end 0.412. This then is our fraction of chlorine molecules that will break apart form thomas chlorine. Thanks for watching. Hope this helps
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Refining petroleum involves cracking large hydrocarbon molecules into smaller, more volatile pieces. A simple example of hydrocarbon cracking is the gas-phase thermal decomposition of butane to give ethane and ethylene: (c) A sample of butane having a pressure of 50 atm is heated at 500 °C in a closed container at constant volume. When equilibrium is reached, what percentage of the butane has been converted to ethane and ethylene? What is the total pressure at equilibrium?

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Textbook Question
At 1000 K, Kp = 2.1 * 106 and ΔH° = - 107.7 kJ for the reaction H21g2 + Br21g2 ∆ 2 HBr1g2. (b) For the equilibrium in part (a), each of the following changes will increase the equilibrium partial pressure of HBr. Choose the change that will cause the greatest increase in the pressure of HBr, and explain your choice. (ii) Adding 0.10 mol of Br2
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Textbook Question
Halogen lamps are ordinary tungsten filament lamps in which the lamp bulb contains a small amount of a halogen (often bromine). At the high temperatures of the lamp, the halogens dissociate and exist as single atoms. (c) When the WBr41g2 diffuses back toward the filament, it decomposes, depositing tungsten back onto the fila- ment. Show quantitatively that the pressure of WBr4 from part (a) will cause the reaction in part (a) to go in reverse direction at 2800 K. [The pressure of Br1g2 is still 0.010 atm.] Thus, tungsten is continually recycled from the walls of the bulb back to the filament, allow-ing the bulb to last longer and burn brighter.
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Textbook Question

The F-F bond in F2 is relatively weak because the lone pairs of electrons on one F atom repel the lone pairs on the other F atom; Kp = 7.83 at 1500 K for the reaction F2(g) ⇌ 2 F(g). (c) Why is the F-F bond in F2 weaker than the Cl-Cl bond in Cl2?

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Textbook Question

The equilibrium constant Kc for the gas-phase thermal decomposition of cyclopropane to propene is 1.0 ⨉105 at 500 K:

(a) What is the value of Kp at 500 K?

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Textbook Question

The equilibrium constant Kc for the gas-phase thermal decomposition of cyclopropane to propene is 1.0 * 105 at 500 K:

(c) Can you alter the ratio of the two concentrations at equilibrium by adding cyclopropane or by decreasing the volume of the container? Explain.

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