Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Chapter 12, Problem 90AE

Sulfuryl chloride (SO2Cl2) decomposes to sulfur dioxide (SO2) and chlorine (Cl2) by reaction in the gas phase. The following pressure data were obtained when a sample containing 5.00 × 10−2 mol sulfuryl chloride was heated to 600. K in a 5.00 × 10−1-L container.

Time (hours): 0.00 1.00 2.00 4.00 8.00 16.00
P SO 2 Cl 2 ( atm ) : 4.93 4.26 3.52 2.53 1.30 0.34

Defining the rate as Δ [ SO 2 Cl 2 ] Δ t ,

a. determine the value of the rate constant for the decomposition of sulfuryl chloride at 600. K.

b. what is the half-life of the reaction?

c. what fraction of the sulfuryl chloride remains after 20.0 h?

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The pressure data obtained when a sample containing 5.00×102mol sulfuryl chloride was heated to 600K in a 5.00×101L container is given. The answers are to be given for each option.

Concept introduction: The change observed in the concentration of a reactant or a product per unit time is known as the rate of the particular reaction. The differential rate law provides the rate of a reaction at specific reaction concentrations.

The half-life of the first order reaction is calculated using the formula,

t12=0.693k

To determine: The value of the rate constant (k) for the decomposition of sulfuryl chloride (SO2Cl2) at 600K

Answer to Problem 90AE

Answer The value of rate constant is 1.68×101s1_ .

Explanation of Solution

Explanation

Given

Temperature is 600K .

Given table,

Time (hours) 0.00 1.00 2.00 4.00 8.00 16.00
PSO2Cl2(atm) 4.93 4.26 3.52 2.53 1.30 0.34

The formula of ideal gas law is,

PV=nRTnV=PRTM=PRT

Where,

  • P is the total pressure.
  • V is the volume.
  • n is the total moles.
  • R is the universal gas constant (0.08206Latm/Kmol) .
  • T is the absolute temperature.
  • M is the concentration.

The concentration of SO2Cl2 is calculated for each given pressure using the above equation. The value of concentration and ln[SO2Cl2] is,

Pressure Concentration of SO2Cl2 (M=PRT) ln[SO2Cl2]
4.93 0.100079 2.3
4.26 0.0865 2.45
3.52 0.0715 2.64
2.53 0.0513 2.97
1.3 0.026 3.65
0.34 0.0069 4.98

The graph is plotted for ln[SO2Cl2] versus time is,

Chemistry, Chapter 12, Problem 90AE

Since, the obtained graph is a straight line, hence, the order of the reaction is first. The slope of graph is 0.168 .

The integral rate law equation of first order reaction is,

ln[SO2Cl2]=kt+ln[SO2Cl2]0 (1)

Where,

  • k is the rate constant.
  • [SO2Cl2]0 is the initial concentration of reactant.
  • t is the time.

The equation (1) is similar to the equation of a straight line, that is,

y=mx+c (2)

Where,

  • y is the y-intercept.
  • x is the x-intercept.
  • m is the slope.
  • c is a constant.

Compare equation (1) and (2).

m=k

Substitute the value of slope in the above equation.

m=kk=(0.168s1)=1.68×101s1_

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The pressure data obtained when a sample containing 5.00×102mol sulfuryl chloride was heated to 600K in a 5.00×101L container is given. The answers are to be given for each option.

Concept introduction: The change observed in the concentration of a reactant or a product per unit time is known as the rate of the particular reaction. The differential rate law provides the rate of a reaction at specific reaction concentrations.

The half-life of the first order reaction is calculated using the formula,

t12=0.693k

To determine: The half life of the given reaction.

Answer to Problem 90AE

Answer

The half life of the given reaction is 4.12s_ .

Explanation of Solution

Explanation

The value of rate constant is 1.68×101s1 .

Formula

The half-life is calculated using the formula,

t12=0.693k

Where,

  • t12 is half life.
  • k is rate constant.

Substitute the values of k in the above equation.

t12=0.693k=0.6931.68×101s1=4.12s_

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The pressure data obtained when a sample containing 5.00×102mol sulfuryl chloride was heated to 600K in a 5.00×101L container is given. The answers are to be given for each option.

Concept introduction: The change observed in the concentration of a reactant or a product per unit time is known as the rate of the particular reaction. The differential rate law provides the rate of a reaction at specific reaction concentrations.

The half-life of the first order reaction is calculated using the formula,

t12=0.693k

To determine: The fraction of sulfuryl chloride (SO2Cl2) left after 20hours .

Answer to Problem 90AE

Answer

The fraction of sulfuryl chloride (SO2Cl2) left after 20hours is 7.4×105mol/L_ .

Explanation of Solution

Explanation

Given

The initial partial pressure of SO2Cl2 at 4.93atm is 0.100079mol/L .

The value of rate constant is 1.68×101s1 .

Time is 20hours .

The conversion of hours (hr) into seconds (s) is done as,

1hr=3600s

Hence,

The conversion of 20hr into seconds is,

20hr=(20×3600)s=72000s

Formula

The rate constant of first order reaction is,

k=2.303tlog([PSO2Cl2]0[PSO2Cl2])

Where,

  • k' is the rate constant.
  • [PSO2Cl2]0 is the initial partial pressure of reactant.
  • [PSO2Cl2] is the fraction of reactant left.
  • t is the time.

Substitute the values of [PSO2Cl2]0,k and t in the above equation.

k=2.303tlog([PSO2Cl2]0[PSO2Cl2])0.168s1=2.30372000slog(0.100079[PSO2Cl2])[PSO2Cl2]=7.4×105mol/L_

Conclusion

Conclusion

The plot of ln[A]vstime graph is a straight line. It means that the given reaction is the first order (a=1) reaction.

The integral rate law equation of first order reaction is,

ln[A]=kt+ln[A]0

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