Chemistry · Ch 6 — Chemical Kinetics

Integrated rate law for gas phase reactions

6.5.7

Integrated rate law for gas phase reactions

Note

The book's printed heading for this subsection reads "6.5.7 Integrated rate law for gas phase f reactions" — a stray italic "f" appears before "reactions" in the print. The intended title, used above, is "Integrated rate law for gas phase reactions".

For the gas phase reaction

\mathrm{A(g) \longrightarrow B(g) + C(g)}

let the initial pressure of A be $P_i$ , which decreases by $x$ within time $t$ .

Pressure of reactant A at time $t$ :

P_A = P_i - x \qquad \text{...(6.11)}

The pressures of the products B and C at time $t$ are

P_B = P_C = x

The total pressure at time $t$ is then

P = P_i - x + x + x = P_i + x

\text{Hence, } x = P - P_i \qquad \text{...(6.12)}

The pressure of A, $P_A$ , at time $t$ is obtained by substitution of Eq. (6.12) into Eq. (6.11). Thus

P_A = P_i - (P - P_i) = P_i - P + P_i = 2P_i - P

The integrated rate law turns out to be

k = \frac{2.303}{t}\,\log_{10}\frac{\mathrm{[A]_0}}{\mathrm{[A]_t}}

with the concentration now expressed in terms of pressures. Thus $\mathrm{[A]_0} = P_i$ and $\mathrm{[A]_t} = P_A = 2P_i - P$ . Substitution gives

k = \frac{2.303}{t}\,\log_{10}\frac{P_i}{2P_i - P} \qquad \text{...(6.13)}

$P$ is the total pressure of the reaction mixture at time $t$ . …

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