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Question
Answer the following in brief.
Derive the integrated rate law for the first-order reaction,
\[\ce{A_{(g)} -> B_{(g)} + C_{(g)}}\] in terms of pressure.
Solution
i. For the gas phase reaction,
\[\ce{A_{(g)} -> B_{(g)} + C_{(g)}}\]
Let initial pressure of A be Pi that decreases by x within time t.
ii. Pressure of reactant A at time t
PA = Pi - x ....(1)
The pressures of the products B and C at time t
PB = PC = x
iii. The total pressure at time t is then
P = Pi - x + x + x = Pi + x
Hence, x = P - Pi …(2)
The pressure of A at time t is obtained by substitution of Eq. (1) into Eq. (2).
Thus
PA = Pi - (P - Pi) = Pi - P + Pi = 2Pi - P
iv. The integrated rate law turns out to be
k = `2.303/"t" "log"_10 ["A"]_0/["A"]_"t"`
The concentration is now expressed in terms of pressures.
Thus, [A]0 = Pi and [A]t = PA = 2Pi - P
Substitution gives in above
`"k" = 2.303/"t" "log"_10 "P"_"i"/(2"P"_"i" - "P")` ...(3)
P is the total pressure of the reaction mixture at time t.
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