Question

the gas-phase decomposition of SO2Cl2,

SO2Cl2(g) ----(arrow)--- SO2(g) + Cl2(g)

the rate constant for the decomposition at 660K is 4.5* 10^-2 s^-1.

a) if we begin with an initial SO2Cl2 pressure of 375 torr, what is the pressure of this substance after 65 s ?
b) At what time will the pressure of SO2Cl2 decline to one-tenth its initial value?

Answer 1

assuming volume of the container of the gas is not changing, and the gas is an ideal gas:
\[\frac{ n_1RT }{ P_1 }=\frac{ n_2RT }{ P_2 } \rightarrow \frac{ n_1 }{ P_1 }=\frac{ n_2 }{ P_2 } \rightarrow \frac{ n_1 }{ n_2 }=\frac{ P_1 }{ P_2 }\]
assuming n1 is the initial number of moles of gas and n2 is the number of moles of gas after 65s:
\[n_2=n_1e^{-4.5\times10^{-2}\times65}\]
by the equation:
\[[A]_t=[A]_0e^{-kt}\]
then:
\[\frac{ n_1 }{ n_1e^{-4.5\times10^{-2}\times65}}=\frac{ 375torr }{ P_2 }\]
\[P_2=375\times e^{-4.5\times10^{-2}\times65}\approx20torr\]

Answer 2

the time the pressure will be one tenth of the initial pressure is given by:
\[\frac{ n_1 }{ n_2 }=\frac{ P_1 }{ 0.1P_1 }=10\]
\[n_2=0.1n_1=n_1e^{-kt} \rightarrow e^{-kt}=0.1\rightarrow kt=-\ln (0.1)=2.3\]
\[t=\frac{ 2.3 }{ k }=\frac{ 2.3 }{ e^{-4.5\times10^{-2}} }=2.4s\]