-->

Chemical Kinetics

Question
CBSEENCH12010206

For the reaction

2NO(g) + Cl2(g) --> 2NOCl(g)

The following data were collected. All the measurements were taken at 263 K:

Experiment No.

Initial [NO] (M)

Initial [Cl2] (M)

Initial rate of disappearance of Cl2(M/min)

1

0.15

0.15

0.60

2

0.15

0.30

1.20

3

0.30

0.15

2.40

4

0.25

0.25

?

 

(a) Write the expression for rate law.

(b) Calculate the value of rate constant and specify its units.

(c) What is the initial rate of disappearance of Cl2 in exp. 4?

Solution
straight a right parenthesis space Rate space law space may space be space written space as
Rate space equals space straight k left square bracket NO right square bracket to the power of straight p space left square bracket Cl subscript 2 right square bracket to the power of straight q
The space initial space rate space becomes
left parenthesis Rate right parenthesis subscript 0 space equals straight K left square bracket NO right square bracket to the power of straight p left square bracket Cl subscript 2 right square bracket to the power of straight q

comparing space experiment space 1 space and space 2
left parenthesis Rate right parenthesis subscript 1 space equals space straight k left parenthesis 0.15 right parenthesis to the power of straight p space left parenthesis 0.15 right parenthesis to the power of straight q space equals 0.60... space left parenthesis 1 right parenthesis
left parenthesis Rate right parenthesis subscript 2 space equals straight k left parenthesis 0.15 right parenthesis to the power of straight p left parenthesis 0.30 right parenthesis to the power of straight q space equals 1.20......... left parenthesis 2 right parenthesis space
Dividing space equation space left parenthesis 2 right parenthesis space by space left parenthesis 1 right parenthesis

fraction numerator left parenthesis Rate right parenthesis subscript 2 over denominator left parenthesis Rate right parenthesis subscript 1 end fraction equals space fraction numerator straight k left parenthesis 0.15 right parenthesis to the power of straight p left parenthesis 0.30 right parenthesis to the power of straight q over denominator straight k left parenthesis 0.15 right parenthesis to the power of straight p left parenthesis 0.15 right parenthesis to the power of straight q end fraction space equals fraction numerator 1.20 over denominator 060 end fraction
Or
2 to the power of straight q space equals space 2 to the power of 1
straight q space equals 1

Order space with space respect space to space Cl subscript 2 equals 1
comparing space experiment space 1 space and space 3
left parenthesis Rate right parenthesis subscript 1 space equals space straight k left parenthesis 0.15 right parenthesis to the power of straight p space left parenthesis 0.15 right parenthesis to the power of straight q space equals 0.60... space left parenthesis 1 right parenthesis
left parenthesis Rate right parenthesis subscript 2 space equals straight k left parenthesis 0.30 right parenthesis to the power of straight p left parenthesis 0.15 right parenthesis to the power of straight q space equals 2.40......... left parenthesis 3 right parenthesis space

Dividing space equation space left parenthesis 2 right parenthesis space by space left parenthesis 1 right parenthesis

fraction numerator left parenthesis Rate right parenthesis subscript 3 over denominator left parenthesis Rate right parenthesis subscript 1 end fraction equals space fraction numerator straight k left parenthesis 0.30 right parenthesis to the power of straight p left parenthesis 0.15 right parenthesis to the power of straight q over denominator straight k left parenthesis 0.15 right parenthesis to the power of straight p left parenthesis 0.15 right parenthesis to the power of straight q end fraction space equals fraction numerator 2.40 over denominator 0.60 end fraction

Or
2 to the power of straight p space equals space 4
2 to the power of straight p space equals 2 squared
straight p equals 2

Thus space order space with space respect space to space NO space is space 2.

straight b right parenthesis space The space rate space law space for space the space reaction space Rate space equals space straight k space left square bracket NO right square bracket squared space left square bracket Cl subscript 2 space right square bracket to the power of 1 space space
Rate space constant space can space be space calculated space by space substituting space the space value space of space rate comma space left square bracket NO right square bracket space and space left square bracket Cl subscript 2 right square bracket space
for space any space experiment
straight k equals fraction numerator Rate over denominator left square bracket NO right square bracket squared left square bracket Cl right square bracket end fraction equals space fraction numerator 0.60 over denominator left parenthesis 0.15 right parenthesis squared left parenthesis 0.15 right parenthesis end fraction

equals fraction numerator 0.60 over denominator 0.003375 end fraction space equals space 177.77 space mol to the power of negative 2 end exponent straight L to the power of negative 2 end exponent space min to the power of minus

straight c right parenthesis space Let space initial space rate space of space disappearance space of space Cl subscript 2 space in space exp. space 4 space space is space straight r subscript 4.
Therefore comma
straight r subscript 4 space equals straight k left square bracket NO right square bracket squared left square bracket Cl subscript 2 right square bracket
equals 177.77 space straight x space left parenthesis 0.25 right parenthesis squared left parenthesis 0.25 right parenthesis
equals 2.78 space straight M divided by min