Question

The following data were obtained during the first order thermal decomposition of $SO_2Cl_2$ at a constant volume.
$SO_2Cl_2(g) → SO_2(g) + Cl_2(g)$

Experiment	Time/s -1	Total pressure/atm
1	0	0.5
2	100	0.6
Calculate the rate of the reaction when total pressure is 0.65 atm.

Answer 1

The thermal decomposition of $SO_2Cl_2$ at a constant volume is represented by the following equation.
$SO_2Cl_2(g) → SO_2(g) + Cl_2(g)$
At t = 0 P0 0 0
At t = 0 P0 - p p p
Pt = (P0 - p) + p + p
After time t, total pressure
⇒ Pt = P0 + p
⇒ p = Pt - P0
Therefore, P0 - P = P0 - (Pt - P0)
= 2P0 - Pt
For a first order reaction,
$k = \frac {2.303}{t} log \ \frac {P_0}{P_0-p}$
$k = \frac {2.303}{t} log \ \frac {P_0}{2P_0-p_t}$
$When \ t = 100\ s$
$k = \frac {2.303}{100\ s} log \ \frac {0.5}{2\times 0.5-0.6}$
$k = 2.231 \times 10^{-3} s^{-1}$
$When\ P_t= 0.65 \ atm$
$P_0+ p= 0.65$
⇒ $p= 0.65 - P_0$
= $0.65 - 0.5$
= $0.15 \ atm$
Therefore, when the total pressure is 0.65 atm, pressure of $SOCl_2$ is
$P_{SOCl_2} = P_0 - p$
= $0.5 - 0.15$
= $0.35\ atm$
Therefore, the rate of equation, when total pressure is 0.65 atm, is given by,
$Rate = k(P_{SOCl_2})$
$Rate = (2.23\times 10^{-3} s^{-1}) (0.35\ atm)$
$Rate = 7.8 \times 10^{- 4}\ atm s^{-1}$