At what time will the pressure of so₂cl₂ decline to 0.50 its initial value? express your answer using two significant figures.

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:

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

In conclusion,

t = ln( p(SO₂Cl₂)₀/p(SO₂Cl₂) )/k

for

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explanation:

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3 hours because it is a decent distance.