The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume.
SO2Cl2(g) → SO2(g) + Cl2(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.
The thermal decomposition of SO2Cl2 at a constant volume is represented by the following equation.
SO2Cl2(g) → SO2(g) + Cl2(g)
At t = 0 P0 0 0
At t = t P0 - p p p
After time, t, total pressure,
Pt = (P0 - p) + p + p
⇒ Pt = P0+ p
⇒ p = Pt - P0
Therefore, P0 - p = P0 -(Pt - P0)
= 2Pt - P0
For a first order reaction,
k = \(\frac{2.303}{t}\)log \(\frac{P_0}{P_0 - p}\)
= \(\frac{2.303}{t}\)log \(\frac{P_0}{2P_0 - P_t}\)
When t = 100 s
k = \(\frac{2.303}{100 \ s}\)log \(\frac{0.5}{2 \times 0.5 - 0.6}\)
= 2.231 × 10-3 s-1
When Pt = 0.65 atm,
P0 + p = 0.65
⇒ p = 0.65 - P0
= 0.65 - 0.5
= 0.15 atm
Therefore, when the total pressure is 0.65 atm, pressure of SO2Cl2 is
PSOCl2 = P0 - 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(PSOCl2)
= (2.23 × 10-3 s-1) (0.35 atm)
= 7.8 × 10-4 atm s-1