SAT Subject Test Chemistry,10 edition

(Marvins-Underground-K-12) #1
is  doubled;    the rate    is  increased   by  a   factor  of  15.9/8.1,   approximately   2.  The rate
expressions of the two trials are:

Dividing    the second  equation    by  the first,

So  r   =   k[A]    [B]^2 , i.e.,   the order   of  the reaction    with    respect to  A   is  1   and with    respect
to B is 2; the overall reaction order is 1 + 2 = 3.
To calculate k, substitute the values from any one of the above trials into the rate
law, e.g.:

Therefore,  the rate    law is  r   =   2.0 M−2 s−1 [A][B]^2.

Note, however, that this result could have been obtained more easily if we had been able to see
without cumbersome substitution that, for example, from trial 1 to trial 2 the rate is quadrupled
when the concentration of B is doubled. Therefore, the rate has to be dependent on the square of
[B].

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