Skip to main content
Ch.15 - Chemical Equilibrium

Chapter 15, Problem 146a

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: (a) Write the equilibrium constant expressions for Kp and Kc.

Verified Solution
Video duration:
2m
This video solution was recommended by our tutors as helpful for the problem above.
320
views
Was this helpful?

Video transcript

Hello everyone today. We have the following problem propane under girls thermal cracking for light oliphant production. One of two thermal decomposition reactions that occur in a reactor is shown below. Right. The equilibrium constant expressions K. P. And K. C. For the reaction. So K. C. And K. P. Are used to express the equilibrium equilibrium constants of gasses and Aquarius mixtures. The only difference is K. Subsea is expressed in Moeller concentrations and K. P. Is expressed in partial pressures and it's the partial pressures of gasses in a closed system in a closed system. So now that we have that out of the way, let's define how we can find Casey into K. P. Well K. C. Is going to be the concentration of our products divided by the concentration of our reactant. K. P. On the other hand, is found by the partial pressure or the pressure of our products divided by the pressure, the partial pressure of our reactant. And so we have our equation here. We can see that we have our products here, our ethane and our ethane and our methane. And then we have our starting product propane as the reactant. And so now we can go ahead and fill in what our case it would be. So our Casey would be the concentration of our products which is going to be our scene times our concentration of methane that's gonna be divided by our reactant which is propane, K. P. On the other hand is going to be the pressure of the products which is going to be our scene times the pressure of methane and then, lastly divided by the partial pressure of our propane. And so these are actually going to be our answers for K. C and for K. P. And with that we have solved the problem overall, I hope that this helped, and until next time.
Related Practice
Textbook Question
Given the Arrhenius equation, k = Ae-Ea>RT, and the rela- tion between the equilibrium constant and the forward and reverse rate constants, Kc = kf>kr, explain why Kc for an exothermic reaction decreases with increasing temperature.
406
views
Textbook Question
Vinegar contains acetic acid, a weak acid that is partially dissociated in aqueous solution: CH3CO2H1aq2 ∆ H+ 1aq2 + CH3CO-1aq2 (b) What is the value of Kc if the extent of dissociation in 1.0 M CH3CO2H is 0.42%?
640
views
Textbook Question
Heavy water, symbolized D2O 1D = 2H2 finds use as a neutron moderator in nuclear reactors. In a mixture with ordinary water, exchange of isotopes occurs according to the following equation: H2O + D2O ∆ 2 HDO Kc = 3.86 at 298 K When 1.00 mol of H2O is combined with 1.00 mol of D2O, what are the equilibrium amounts of H2O, D2O, and HDO (in moles) at 298 K? Assume the density of the mixture is constant at 1.05 g>cm3.
285
views
Textbook Question

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?

329
views
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
486
views
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.
328
views