Using the given data, determine the rate constant of this reaction. A+B ——> C+D

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  1. Preliminary(Trial) A(M) B(M) Rate (M/s)
    1. 0.250 0.200 0.0215
    2 0.250 0.520 0.145
    3 0.350 0.200 0.0301
    K=?
    What units ought to be utilized?
    Request of a Reaction:
    Request of a response is the whole of the terms raised to the intensity of the focus species in the articulation characterizing the rate law of a response.
    Leave the alone of the response be:
    Rate=k(A)a(Bb)
    where An and B are the two reactants.
    At that point, the request for the response is (a+b)
    Request concerning A will be an and request as for B will be b.
    Request of a response tells the reliance of the response rate on the convergence of various responding species.
    Answer and Explanation:
    Let the rate law for the response be
    Rate=k(A)a(Bb)
    For preliminary 1,
    0.0215=k(0.250)a(0.200)b
    For preliminary 2,
    0.145=k(0.250)a(0.520)b
    For preliminary 3,
    0.0301=k(0.350)a(0.200)b
    Isolating the rate laws for preliminary 2 by preliminary 1:
    0.0215=0.250a×0.200b/0.145=0.250a×0.520b
    Dropping the basic terms gives the estimation of b:
    0.148=0.384b
    Taking log the two sides:
    log0.148=log(0.384b)
    Since, logxy=ylogx
    −0.8297=b(−0.4156)
    The estimation of b is therefore:
    b=2
    Correspondingly, separating the rate law in preliminary 3 by preliminary 1 gives the estimation of a:
    0.0215=0.250a×0.200b/0.0301=0.350a×0.200b
    0.714=0.714a
    Therefore, a = 1
    Therefore, the estimation of k can be determined by utilizing rate law articulation for any one preliminary:
    0.215(M/s)=k×0.2501(M)×0.2002(M2)
    ⇒k=21.5M−2s−1

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