Question

At what time will the pressure of SO₂Cl₂ decline to 0.50 its initial value? Express your answer using two significant figures.

Answer 1

Answer: The time is 0.69/k seconds

Step-by-step explanation:

The following integrated first order rate law

ln[SO₂Cl₂] - ln[SO₂Cl₂]₀ = - k×t

where

[SO₂Cl₂] concentration at time t,

[SO₂Cl₂]₀ initial concentration,

k rate constant

Therefore, the time elapsed after a certain concentration variation is given by:

`t=(ln[SO_(2)Cl_(2)]_(0) - ln[SO_(2)Cl_(2)])/(k)=(ln([SO_(2)Cl_(2)]_(0))/([SO_(2)Cl_(2)]) )/(k)`

We could assume that SO₂Cl₂ behaves as a ideal gas mixture so partial pressure is proportional to concentration:

`p_{(SO_(2)Cl_(2))}V = n_{(SO_(2)Cl_(2))}RT`

`[SO_(2)Cl_(2)]= \frac{n_{(SO_(2)Cl_(2))}}{V}}=\frac{p_{(SO_(2)Cl_(2))}}{RT}}`

In conclusion,

t = ln( p(SO₂Cl₂)₀/p(SO₂Cl₂) )/k

`t=\frac{ln\frac{p_{(SO_(2)Cl_(2))}_(0)}{p_{(SO_(2)Cl_(2))}} }{k}`

for

`p_{(SO_(2)Cl_(2))}=0.5p_{(SO_(2)Cl_(2))}_(0)`

`t=\frac{ln\frac{p_{(SO_(2)Cl_(2))}_(0)}{0.5p_{(SO_(2)Cl_(2))_(0)}} }{k}`

`t=(ln(1)/(0.5) )/(k)`

`t=(ln(2))/(k)`

`t=(0.69)/(k)}`