J.L. Carriker et al. / Journal of Molecular Catalysis A: Chemical 267 (2007) 218–223
223
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−d[D2]
= k1[D2] + k2[D2].
dt
water is present in such large excess that it is assumed to have a
constant concentration over the time of the reaction. Integration
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1
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= k1[D2] − k1[HD].
dt
2
Substitution of Eq. (8) for [D2] results in
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d[HD]
1
2
= k1[D2]0e−(k +k )t
−
k1[HD],
1
2
dt
which is in the form
dx
= ae−bt − cx
dt
where a = k1[D2]0, b = k1 + k2, c = 1/2k1, and x = [HD]t.
The solution to the above differential equation is
a
xt =
e−bt + Ke−ct
c − b
where K is the constant of integration.
The constant of integration is determined by using the fact
that at t = 0, [HD] = 0, thus:
a
a
K = −
,
resulting in : xt =
e−bt
−
e−ct
.
(b) P.S. Wagenknecht, E.J. Sambriski, in: S.G. Pandalai (Ed.), Inorganic
Chemistry, Transworld Research Network, Kerala, India, 2003, pp. 33–50.
c − b
c − b
c − b
´
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Back substitution for a, b, c, and x results in Eq. (9). Finally,
mass balance gives Eq. (10).
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