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Chemical Kinetics

Question
CBSEENCH12010113
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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
2

0
100

0.4
0.7

 Calculate the rate constant. (Given :  log 4 = 0.6021, log 2 = 0.3010)

Solution

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 SO subscript 2 Cl subscript 2 left parenthesis straight g right parenthesis space rightwards arrow space space SO subscript 2 left parenthesis straight g right parenthesis space space plus Cl subscript 2 space left parenthesis straight g right parenthesis At space straight t space equals space 0 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 0 space space space space space space space space space space space space space space 0 At space straight t equals space space straight t 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 straight p 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 comma straight P subscript straight i 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 space plus straight p Which space space on space rearrangement space gives colon straight p equals space straight P subscript straight t minus straight P subscript 0 Therefore comma 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 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 equals fraction numerator space 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 space 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 i end fraction When space straight t space equals space 100 straight s straight k equals fraction numerator 2.303 over denominator 100 end fraction log space fraction numerator 0.4 over denominator 2 space straight x space 0.4 space minus 0.7 end fraction space equals 1.38 space straight x space 10 to the power of negative 2 end exponent space straight s to the power of negative 1 end exponent

Some More Questions From Chemical Kinetics Chapter

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