General Chemistry
General Chemistry
4th Edition
ISBN: 9781891389603
Author: Donald A. McQuarrie, Peter A. Rock, Ethan B. Gallogly
Publisher: University Science Books
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Chapter 19, Problem 19.81P
Interpretation Introduction

Interpretation:

The total pressure after equilibrium is reestablished for the given reaction has to be calculated.

  CO(g)+Cl2(g)COCl2(g)

Expert Solution & Answer
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Explanation of Solution

The reaction is given below.

  CO(g)+Cl2(g)COCl2(g)KP=PCOCl2(PCO)(PCl2)

The first set of equilibrium partial pressures are PCO=PCl2=1.09barandPCOCl2=0.144bar.

By substituting these values in the above expression, the value of equilibrium constant can be calculated.

  KP=0.144bar(1.09bar)(1.09bar)=0.1212bar1

The pressure of CO(g) immediately after addition of CO(g) can be calculated from the relation for the total pressure.

  Ptotal=PCO+PCl2+PCOCl23.31bar=PCO+1.09bar+0.144barPCO=2.076bar

According to Le Chatelier’s principle, the position of the equilibrium will shift towards right in response to the addition of CO(g).  Thus, some of CO(g) and Cl2(g) will decompose to produce COCl2(g).  Let it be x.

To determine the new equilibrium conditions, a pressure table can be set up as given below.

  Pressure(bar)CO(g)+Cl2(g)COCl2(g)Initial2.0761.090.144Changex+x+xEquilibrium2.076x1.09x0.144+x

Substitute the equilibrium pressures into the equilibrium – constant expression as shown below.

  KP=0.144bar+x(2.076barx)(1.09barx)=0.1212bar10.144bar+x=(2.076barx)(1.09barx)(0.1212bar1)(0.1212bar1)x21.3837x+0.1299bar=0x=11.32bar,0.0946bar

The value 11.32bar results into negative value of concentration, thus this must be rejected.  Therefore, the value of x is 0.0946bar.

Then,

  Pressure(bar)CO(g)+Cl2(g)COCl2(g)Initial2.0761.090.144Change0.09460.0946+0.0946Equilibrium1.98140.99540.2386

The total pressure after equilibrium is re-established can be calculated as given below.

  Ptotal=PCO+PCl2+PCOCl2=1.9814bar+0.9954bar+0.2386bar=3.2154bar3.22bar

As a final check, it can be shown that,

  PCOCl2(PCO)(PCl2)=0.2386bar(1.9814bar)(0.9954bar)=0.121bar1

which agrees with KP=0.1212bar1.

Therefore, the total pressure after equilibrium is re-established is 3.22bar.

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Chapter 19 Solutions

General Chemistry

Ch. 19 - Prob. 19.11PCh. 19 - Prob. 19.12PCh. 19 - Prob. 19.13PCh. 19 - Prob. 19.14PCh. 19 - Prob. 19.15PCh. 19 - Prob. 19.16PCh. 19 - Prob. 19.17PCh. 19 - Prob. 19.18PCh. 19 - Prob. 19.19PCh. 19 - Prob. 19.20PCh. 19 - Prob. 19.21PCh. 19 - Prob. 19.22PCh. 19 - Prob. 19.23PCh. 19 - Prob. 19.24PCh. 19 - Prob. 19.25PCh. 19 - Prob. 19.27PCh. 19 - Prob. 19.28PCh. 19 - Prob. 19.29PCh. 19 - Prob. 19.30PCh. 19 - Prob. 19.31PCh. 19 - Prob. 19.32PCh. 19 - Prob. 19.33PCh. 19 - Prob. 19.34PCh. 19 - Prob. 19.35PCh. 19 - Prob. 19.36PCh. 19 - Prob. 19.37PCh. 19 - Prob. 19.38PCh. 19 - Prob. 19.39PCh. 19 - Prob. 19.40PCh. 19 - Prob. 19.43PCh. 19 - Prob. 19.44PCh. 19 - Prob. 19.45PCh. 19 - Prob. 19.46PCh. 19 - Prob. 19.47PCh. 19 - Prob. 19.48PCh. 19 - Prob. 19.49PCh. 19 - Prob. 19.50PCh. 19 - Prob. 19.51PCh. 19 - Prob. 19.52PCh. 19 - Prob. 19.53PCh. 19 - Prob. 19.54PCh. 19 - Prob. 19.55PCh. 19 - Prob. 19.56PCh. 19 - Prob. 19.57PCh. 19 - Prob. 19.58PCh. 19 - Prob. 19.59PCh. 19 - Prob. 19.60PCh. 19 - Prob. 19.61PCh. 19 - Prob. 19.62PCh. 19 - Prob. 19.63PCh. 19 - Prob. 19.64PCh. 19 - Prob. 19.65PCh. 19 - Prob. 19.66PCh. 19 - Prob. 19.67PCh. 19 - Prob. 19.68PCh. 19 - Prob. 19.69PCh. 19 - Prob. 19.70PCh. 19 - Prob. 19.71PCh. 19 - Prob. 19.72PCh. 19 - Prob. 19.73PCh. 19 - Prob. 19.74PCh. 19 - Prob. 19.75PCh. 19 - Prob. 19.76PCh. 19 - Prob. 19.77PCh. 19 - Prob. 19.78PCh. 19 - Prob. 19.79PCh. 19 - Prob. 19.80PCh. 19 - Prob. 19.81PCh. 19 - Prob. 19.82PCh. 19 - Prob. 19.83PCh. 19 - Prob. 19.84PCh. 19 - Prob. 19.86PCh. 19 - Prob. 19.91PCh. 19 - Prob. 19.92PCh. 19 - Prob. 19.93PCh. 19 - Prob. 19.94P
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