The following data were obtained during the first order thermal decomposition of SO_{2}Cl_{2} at a constant volume :SO_{2}Cl_{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.

#### Solution

The thermal decomposition of SO_{2}Cl_{2} at a constant volume is represented by the following equation.

SO_{2}Cl_{2} (g) → SO_{2} (g) + Cl_{2} (g)

At t = 0 | `P_0` | 0 | 0 |

At t = t | `P_0-p` | p | p |

After time, *t*, total pressure, `P_t = (P_0 - p) + p + p`

`=>P_t = P_0 + p`

`=>p = P_t - P_0`

Therefore `P_0 - p = P_0 - (P_t-P_0)`

= 2 P_{0} − P_{t}

For a first order reaction,

`k = 2.303/t log P_0/(P_0-p)`

When t = 100 s, `k = 2.303/100s log 0.5/(2xx0.5-0.6)`

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`))

= (2.23 × 10^{−3} s^{−1}) (0.35 atm)

= 7.8 × 10^{−4} atm s^{−1}