Answer: The time is 0.69/k seconds
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:
[tex]t=\frac{ln[SO_{2}Cl_{2}]_{0} - ln[SO_{2}Cl_{2}]}{k}=\frac{ln\frac{[SO_{2}Cl_{2}]_{0}}{[SO_{2}Cl_{2}]} }{k}[/tex]
We could assume that SO₂Cl₂ behaves as a ideal gas mixture so partial pressure is proportional to concentration:
[tex]p_{(SO_{2}Cl_{2})}V = n_{(SO_{2}Cl_{2})}RT[/tex]
[tex][SO_{2}Cl_{2}]= \frac{n_{(SO_{2}Cl_{2})}}{V}}=\frac{p_{(SO_{2}Cl_{2})}}{RT}}[/tex]
In conclusion,
t = ln( p(SO₂Cl₂)₀/p(SO₂Cl₂) )/k
[tex]t=\frac{ln\frac{p_{(SO_{2}Cl_{2})}_{0}}{p_{(SO_{2}Cl_{2})}} }{k}[/tex]
for
[tex]p_{(SO_{2}Cl_{2})}=0.5p_{(SO_{2}Cl_{2})}_{0}[/tex]
[tex]t=\frac{ln\frac{p_{(SO_{2}Cl_{2})}_{0}}{0.5p_{(SO_{2}Cl_{2})_{0}}} }{k}[/tex]
[tex]t=\frac{ln\frac{1}{0.5} }{k}[/tex]
[tex]t=\frac{ln(2)}{k}[/tex]
[tex]t=\frac{0.69}{k}}[/tex]
