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

Question

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## Answer ( 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