Calculating Equilibrium Concentrations for a Gas Equilibrium Problem

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Discussion Overview

The discussion revolves around calculating the equilibrium concentrations of gases in a chemical equilibrium involving PCl5, PCl3, and Cl2. The scenario involves determining the concentrations after a certain percentage of PCl5 has decomposed in a closed container.

Discussion Character

  • Homework-related

Main Points Raised

  • One participant presents the initial problem, stating that PCl5 is 45% decomposed and asks for the equilibrium concentrations of PCl5, PCl3, and Cl2 in a 1.00 L container.
  • Another participant calculates the remaining amount of PCl5 after decomposition and questions how much PCl5 reacted to determine the amounts of products formed.
  • A subsequent reply suggests that the amounts of PCl3 and Cl2 formed are equal to the amount of PCl5 that decomposed.
  • A later response confirms the previous calculation regarding the amounts of PCl3 and Cl2.

Areas of Agreement / Disagreement

Participants appear to agree on the calculations regarding the amounts of PCl3 and Cl2 formed from the decomposition of PCl5, but there is no explicit consensus on the overall approach to solving the problem.

Contextual Notes

The discussion does not clarify the assumptions regarding the reaction conditions or the specific definitions of equilibrium concentrations used in the calculations.

Who May Find This Useful

Students or individuals studying chemical equilibrium, particularly those interested in gas-phase reactions and equilibrium calculations.

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PCl5 <---> PCl3 + Cl2
all are gases

PCl5 is 45% decomposed
if .110 mol of PCl5 is put in a 1.00 L container, what will
be the equilibrium concentrations of PCl5, PCl3, and Cl2?

PCl5 = .110 * .45 = .0495
.110-.0495= .0605 mol/L

so I've got .0605 mol/L for all 3 since all have the
same coeff. Is this how to solve?
 
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0.110 - 0.0495 = 0.0605

So there's 0.0605 mol of PCl5, but how much PCl5 reacted? Therefore how much product was formed?
 
.0495 mol of PCl3
.0495 mol of Cl2

right?
 
You got it.
 

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