(a)
Interpretation:
The value of the rate constant for the decomposition of sulfuryl chloride at 600 K should be determined.
Concept Introduction:
Integrated rate laws for zero, first and second order reactions are,
Zeroth order:
First order:
Second order:
(a)
Answer to Problem 106AE
Explanation of Solution
If
Time(hour) | 0.00 | 1.00 | 2.00 | 4.00 | 8.00 | 16.00 |
Ptotal(atm) | 4.93 | 5.60 | 6.34 | 7.33 | 8.56 | 9.52 |
PSO2Cl2 | 4.93 | 4.26 | 3.52 | 2.53 | 1.30 | 0.34 |
Ln PSO2Cl2 | 1.595 | 1.449 | 1.258 | 0.928 | 0.262 | -1.08 |
Pressure of gas is directly proportional to concentration.
The graph of ln PSO2Cl2 versus time is linear. So, the reaction of decomposition of SO2Cl2 is first order.
Integrated rate law for the reaction is,
The slope of the graph gives the value for rate constant.
(b)
Interpretation:
The half-life of the reaction should be calculated.
Concept Introduction:
Half-life for first order reaction can be calculated by,
(b)
Answer to Problem 106AE
Explanation of Solution
(c)
Interpretation:
The pressure in the vessel after 0.500 h and after 12.0 h should be calculated.
Concept Introduction:
Integrated rate law for the first order reaction;
[A]t − concentration of A at time t
[A]0 − initial concentration of A
k − rate constant
t − time
(c)
Answer to Problem 106AE
After 0.500 h = 5.33 atm
After 12.00 h = 9.20 atm
Explanation of Solution
After 0.500 h;
After 12.00 h;
(d)
Interpretation:
The fraction of the sulfuryl chloride remains after 20.0 h should be calculated.
Concept Introduction:
Integrated rate law for the first order reaction;
[A]t − concentration of A at time t
[A]0 − initial concentration of A
k − rate constant
t − time
(d)
Answer to Problem 106AE
Fraction of SO2Cl2 left =
Explanation of Solution
Fraction of SO2Cl2 left =
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Chapter 15 Solutions
Chemical Principles
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