## 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.
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