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
P_t = (P₀ - p) + p + p
After time t, total pressure
⇒ P_t = P₀ + p
⇒ p = P_t - P₀
Therefore, P₀ - P = P₀ - (P_t - P₀)
= 2P₀ - P_t
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,
\(Rate = k(P_{SOCl_2})\)
\(Rate = (2.23\times 10^{-3} s^{-1}) (0.35\  atm)\)
\(Rate = 7.8 \times 10^{- 4}\  atm s^{-1}\)