Home
Class 12
CHEMISTRY
The rate of formation of a dimer in a se...

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.

Text Solution

Verified by Experts

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)`
Promotional Banner

Topper's Solved these Questions

  • SAMPLE PAPER - 18 (UNSOLVED)

    FULL MARKS|Exercise PART - III|5 Videos

  • SAMPLE PAPER - 18 (UNSOLVED)

    FULL MARKS|Exercise PART - IV|5 Videos

  • SAMPLE PAPER - 18 (UNSOLVED)

    FULL MARKS|Exercise PART - IV|5 Videos

  • SAMPLE PAPER - 17 (UNSOLVED)

    FULL MARKS|Exercise PART - IV|2 Videos

  • SAMPLE PAPER - 19 ( UNSOLVED)

    FULL MARKS|Exercise PART - IV|4 Videos

Similar Questions

Explore conceptually related problems

The rate constant for a first order reaction is 1.54xx10^(-3)s^(-1) . Calculate its half life time.

The initial rate of a first order reaction is 5.2xx10^(-6) "mol lit"^(-1)s^(-1) 298 K. When the initial concentration of reactant is 2.6xx10^(-3)"mol lit"^(-1) , calculate the first order rate constant of the reaction at the same temperature.

The rate constant of a first order reaction is 2.0xx10^(-6)s^(-1) . The initial concentration of the reactant is 0.10 "mol dm"^(-3) . What is the value of the initial rate ?