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    Determining the Rate Equation

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    Into a thoroughly washed and dried Erlenmeyer flask, place the required amount of H2C2O4 and any distilled water. The amounts are dictated by the experiment that you are doing.

    Reagents exp.1 exp. 2 exp.3
    H2C2O4 20.00ml 20.00ml 10.00ml
    KMnO4 10.00ml 5.00ml 10.00ml
    H2O 0.00 ml 5.00ml 10.00ml

    Into a 15cm test tube place the required amount of KMnO4.

    Add the permanganate to the oxalic acid and commence timing when you have emptied the permanganate tube. Mix thoroughly buy swirling the Erlenmeyer flask and continue swirling until the solution turns a light yellow/brown color. Stop timing and record the time it actually took for the reaction to take place.

    Repeat this with a second and third trial. Take the average of these three as the reaction time.

    Repeat steps 3 through 5 for experiments 2 and 3.

    These were the times I recorded in the lab for each of the three experiments in minutes.

    Exp1 Exp2 Exp 3
    3.07 4.04 5.20
    3.10 4.19 5.18
    3.04 4.15 5.24

    Determine the rate for each of the three experiments. Remember this is just the [KMnO4]/taverage .

    Write-out the full rate equation for each experiment. Again remember that this is equal to:
    RateExp#x = k[KMnO4]x[H2C2O4]y. You now have three equations with three unknowns (k,x,y), use the procedure outlined in the introduction to determine the values of these unknowns.

    © BrainMass Inc. brainmass.com October 9, 2019, 3:53 pm ad1c9bdddf
    https://brainmass.com/chemistry/chemical-kinetics/determining-the-rate-equation-16266

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    Dear student,
    This experiment is very simple. In this experiment you have to estimate the values of x,y and k, which are defined in your write-up.

    For experiment 1 average time of reaction:
    t1average = (3.07+3.10+3.04)/3 = 3.07 min = 3.07*60 = 184.20 sec
    similarily,
    t2average = (4.04+4.19+4.15)/3 = 12.38/3 min = 12.38*60/3 = 247.6 sec

    t3average = (5.20+5.18+5.24)/3 = 15.62/3 min = 15.62*60/3 = 312.4 sec

    I hope you have used 0.02M-KMnO4 and 0.5M-H2C2O4.
    rate R = [KMnO4]/taverage = k*[KMnO4]^x *[H2C2O4]^y
    Take ln both sides:
    ln([KMnO4]/taverage) = ln(k) + x*ln([KMnO4]) + y*ln([H2C2O4])
    let ln(k) = K
    ln([KMnO4]/taverage) = K ...

    Solution Summary

    This experiment is very simple. In this experiment you have to estimate the values of x,y and k, which are defined in your write-up. For experiment 1: rate R1 = [KMnO4]/t1average = k*[KMnO4]^x *[H2C2O4]^y.

    $2.19