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The rate of formation of a dimer in a second order dimerisation reaction is 9*5xx10^(-5)molL^(-1)s^(-1)" at "0*01molL^(-1) monomer concentration. Calculate the rate constant. |
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Answer» Solution :If the monomer is represented by A, then the reaction is : `2Ato(A)_(2)` As the reaction is of SECOND ORDER, the rate of reaction will be given by : Rate `=k[A]^(2)` Rate of reaction `=-(1)/(2)(d[A])/(dt)=+(d[A_(2)])/(dt)=9*5xx10^(-5)molL^(-1)s^(-1)` `:.9*5xx10^(-5)"MOL "L^(-1)s^(-1)=k(0*01"mol "L^(-1))^(2)=kxx10^(-4)("mol "L^(-1))^(2)" or "k=0*95L" mol"^(-1)s^(-1)` |
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