Chemical Kinetics

Chemical Kinetics

Question

The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume :
SO2Cl2 (g) → SO2 (g) + Cl2 (g)

Experiment

Time/s−1

Total pressure/atm

1

0

0·4

2

100

0·7

Calculate the rate constant.
(Given : log 4 = 0·6021, log 2 = 0·3010)

Answer

The thermal decomposition of SO2Cl2 at a constant volume is represented by the following equation:
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space SO subscript 2 Cl subscript 2 left parenthesis straight g right parenthesis space space rightwards arrow space space space SO subscript 2 left parenthesis straight g right parenthesis space space plus space space space space Cl subscript 2 left parenthesis straight g right parenthesis
At space space space straight t equals space space 0 space space space space space space space space space space space space space straight P subscript 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space space 0 space space space space space space space space space space space space space space space space space space space 0
At space space space straight t equals space straight t space space space space space space space space space space space space space space space straight P subscript 0 minus straight p space space space space space space space space space space space space space space space space space space space space space space space straight p space space space space space space space space space space space space space space space space space space straight p

After space time space straight t comma space total space pressure space us space given space as comma
straight P subscript straight t space equals space left parenthesis straight P subscript 0 minus straight p right parenthesis space plus straight p plus straight p
straight P subscript straight t space equals space straight P subscript 0 plus straight p
which space on space rearangement space gives colon space
straight p equals straight P subscript straight t minus straight P subscript 0
therefore
straight P subscript 0 minus straight p space equals straight P subscript 0 minus left parenthesis straight P subscript straight t minus straight P subscript 0 right parenthesis space equals 2 straight P subscript 0 minus straight P subscript straight t

For space straight a space first space order space reaction comma

straight k space equals fraction numerator 2.303 over denominator straight t end fraction log space fraction numerator straight P subscript 0 over denominator straight P subscript 0 minus straight p end fraction
equals fraction numerator 2.303 over denominator straight t end fraction log space fraction numerator straight P subscript 0 over denominator 2 straight P subscript 0 minus straight P subscript straight t end fraction
equals space straight t equals 100 straight s
straight k equals fraction numerator 2.303 space space over denominator 100 end fraction log space fraction numerator 0.42 over denominator 0.4 minus 0.7 end fraction space equals 1.386 space straight x space 10 to the power of negative 2 end exponent straight s to the power of negative 1 end exponent

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