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The rate of formation of a dimer in a second order reaction is 7.5xx10^(-3)"mol L"^(-1)s^(-1) at 0.05"mol L"^(-1) monomer concentration. Calculate the rate constant. |
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Answer» SOLUTION :Let US consider the dimerisation of a monomer M `2Mrarr(M)_(2)` `"Rate "=k[M]^(n)` `"Given that n = 2 and [M]"="0.05 MOL L"^(-1)` `"Rate "=7.5xx10^(-3)" mol L"^(-1)s^(-1)` `k=("Rate")/([M]^(n))RARR k=(7.5xx10^(-3))/((0.05)^(2))="3 mol"^(-1)L s"^(-1)` |
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