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 | Total pressure/atm |

1 | 0 | 0.5 |

2 | 100 | 0.6 |

Calculate the rate of the reaction when total pressure is 0.65 atm

When t = 0, the total partial pressure is P_{0} = 0.5 atm

When time t = t, the total partial pressure is P_{t} = P_{0} + p

P_{0}-p = P_{t}-2p, but by the above equation, we know p = P_{t}-P_{0}

Hence, P_{0}-p = P_{t}-2(P_{t}-P_{0})

Thus, P_{0}-p = 2P_{0} – P_{t}

We know that time

Where, k- rate constant

[R]_{°} -Initial concentration of reactant

[R]-Concentration of reactant at time ‘t’

Here concentration can be replaced by the corresponding partial pressures.

Hence, the equation becomes,

⇒ → equation 1

At time t = 100 s, P_{t} = 0.6 atm and P_{0} = 0.5 atm,

Substituting in equation 1,

⇒

Thus, k = 2.231 × 10^{-3} s^{-1}

The rate of reaction R = k × P_{S02Cl2}

When total pressure P_{t} = 0.65 atm and P_{0} = 0.5 atm, then

P_{S02Cl2} = 2P_{0}-P_{t}

Thus, substituting the values, P_{S02Cl2} = 2(0.5)-0.6 = 0.35 atm

R = k × P_{S02Cl2} = 2.231 × 10^{-3} s^{–1} × 0.35

Rate of the reaction R = 7.8 × 10^{-4}atm s^{–1}

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