PCI5 (2) = PCI3 (g) + Cl2 (g) AH = Positive Given that when 8.4 mol of PCI5 (g) is mixed with 1.8 mol of PCI3 (g) and allowed to come to equilibrium in a 10 dm³ container the amount of PCI5 (g) at equilibrium is 7.2 mol.

PCI5 (2) = PCI3 (g) + Cl2 (g) AH = Positive Given that when 8.4 mol of PCI5 (g) is mixed with 1.8 mol of PCI3 (g) and allowed to come to equilibrium in a 10 dm³ container the amount of PCI5 (g) at equilibrium is 7.2 mol.

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter19: The Chemistry Of The Main-group Elements

Section: Chapter Questions

Problem 60QRT

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Concept explainers

Solutions

Solutions can be considered as homogeneous mixtures which contain two or more than two chemical substances. The chemical substance which is the greater part is defined as solvent. And the chemical substance that is dissolved in a solvent is referred to as solute. Solution chemistry plays an important role in chemistry since it supports many commercial applications of solutions.

Question

4 (a) Calculate the value of Kc for the reaction:
PCI5 (g)
PC13 (g)
Cl2 (g)
AH = Positive
Given that when 8.4 mol of PCI5 (g) is mixed with 1.8 mol of PCI3 (g) and allowed to come to
equilibrium in a 10 dm³ container the amount of PCI5 (g) at equilibrium is 7.2 mol.
Kc =
(b) Explain the effect of the following changes below on the value of Kc:
(i) Increasing temperature
(ii) Lowering the concentration of chlorine (Cl2)
(iii) Addition of a catalyst

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