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.

Answer 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, P_t = (P₀ − p) + p + p

⇒ P_t = (P₀ + p)

⇒ p = P_t − P₀

∴ P₀ − p = P₀ − (P_t − P₀)

= 2 P₀ − P_t

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 P_t = 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

Answer 2

If P₀ is the initial pressure and P_t at time t, we have

k = `2.303/t log_10  P_0/(2P_0 - P_t)`

Given, P₀ = 0.5 atm, P_t (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, P_t = 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