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Question
The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume.
\[\ce{SO2Cl2_{(g)} -> SO2_{(g)} + Cl2_{(g)}}\]
| 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.
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Solution 1
The thermal decomposition of SO2Cl2 at a constant volume is represented by the following equation.
\[\ce{SO2Cl2_{(g)} -> SO2_{(g)} + Cl2_{(g)}}\]
| 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
∴ P0 − p = P0 − (Pt − P0)
= 2 P0 − Pt
For a first order reaction,
k = `2.303/t log P_0/(P_0 - p)`
When t = 100 s, k = `2.303/(100 s)log 0.5/(2 xx 0.5 - 0.6)`
When Pt = 0.65 atm,
P0 + p = 0.65
⇒ p = 0.65 − P0
= 0.65 − 0.5
= 0.15 atm
∴ When the total pressure is 0.65 atm, pressure of SOCl2 is \[\ce{P_{SO_2Cl_2}}\] = P0 − p
= 0.5 − 0.15
= 0.35 atm
∴ The rate of equation, when total pressure is 0.65 atm, is given by,
Rate = \[\ce{k(P_{SO_2Cl_2})}\]
= (2.23 × 10−3 s−1) × (0.35 atm)
= 7.8 × 10−4 atm s−1
Solution 2
If P0 is the initial pressure and Pt at time t, we have
k = `2.303/t log_10 P_0/(2P_0 - P_t)`
Given, P0 = 0.5 atm, Pt (at t = 100 s) = 0.6 atm
k = `2.303/t log_10 P_0/(2P_0 - P_t)`
= `2.303/100 log_10 0.5/(2 xx 0.5 - 0.6)`
= 2.2318 × 10−3 s−1
When, Pt = 0.65 atm, i.e., P0 + x = 0.65
∴ x = 0.65 − P0
= 0.65 − 0.5
x = 0.15 atm
Pressure of SO2Cl2 when the total pressure is 0.65 atm, i.e.,
`P_(SO_2Cl_2)` = P0 − x
= 0.50 − 0.15
= 0.35 atm
∴ The required rate of reaction = k × `P_(SO_2Cl_2)`
= 2.2318 × 10−3 × 0.35
= 7.81 × 10−4 atm s−1
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