Ch.15 - Chemical Equilibrium

McMurry8th EditionChemistryISBN: 9781292336145Non è quello che usi tu?Cambia libro di testo

McMurry 8th Edition

Ch.15 - Chemical Equilibrium

Problema 146c

Capitolo 15, Problema 146c

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?

Guida verificata passo dopo passo

Write the balanced chemical equation for the decomposition of butane: \[ \text{C}_4\text{H}_{10(g)} \rightarrow \text{C}_2\text{H}_{6(g)} + \text{C}_2\text{H}_{4(g)} \]

Define the initial conditions: Initial pressure of butane is 50 atm, and the initial pressures of ethane and ethylene are 0 atm.

Set up an ICE (Initial, Change, Equilibrium) table to track the changes in pressure for each species. Let 'x' be the change in pressure for butane that decomposes.

Express the equilibrium pressures in terms of 'x': \[ P_{\text{C}_4\text{H}_{10}} = 50 - x \], \[ P_{\text{C}_2\text{H}_6} = x \], \[ P_{\text{C}_2\text{H}_4} = x \].

Calculate the total pressure at equilibrium: \[ P_{\text{total}} = (50 - x) + x + x = 50 + x \].

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.

Durata del video:

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

Hydrocarbon Cracking

Hydrocarbon cracking is a chemical process that breaks down large hydrocarbon molecules into smaller, more useful ones, typically through heat or catalysts. This process is essential in petroleum refining, as it increases the yield of lighter hydrocarbons like ethane and ethylene, which are valuable for fuel and chemical production. Understanding the mechanisms and products of cracking is crucial for predicting the outcomes of reactions involving hydrocarbons.

Video consigliato:

Percorso guidato

04:29

Intro To Hydrocarbons

Equilibrium in Chemical Reactions

Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, resulting in constant concentrations of reactants and products over time. In the context of the butane cracking reaction, reaching equilibrium means that the amounts of butane, ethane, and ethylene will stabilize, allowing for the calculation of conversion percentages and total pressure. The concept of equilibrium is fundamental in predicting the behavior of reactions under specific conditions.

Video consigliato:

Percorso guidato

04:21

Chemical Equilibrium Concepts

Gas Laws and Pressure Calculations

Gas laws describe the relationships between pressure, volume, temperature, and the number of moles of a gas. In this scenario, the ideal gas law (PV=nRT) can be applied to determine the total pressure at equilibrium after the reaction. Understanding how to manipulate these variables is essential for calculating the changes in pressure and volume that occur during the cracking process, especially in a closed system.

Video consigliato:

Percorso guidato

01:43

Combined Gas Law

Pratica correlata

Domanda del libro di testo

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 K_p and K_c.

322

views

Domanda del libro di testo

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.

329

views

Domanda del libro di testo

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

Domanda del libro di testo

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.