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

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:

$t=\frac{ln[SO_{2}Cl_{2}]_{0} - ln[SO_{2}Cl_{2}]}{k}=\frac{ln\frac{[SO_{2}Cl_{2}]_{0}}{[SO_{2}Cl_{2}]} }{k}$

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

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

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

$t=\frac{ln\frac{1}{0.5} }{k}$

$t=\frac{ln(2)}{k}$

$t=\frac{0.69}{k}}$