Question

If the initial concentration of so2cl2 is 0.150 m , what is the concentration of so2cl2 after 210 s ?

Answer 1

Answer:

Explanation:

Use integrated first order rate law, which give you the concentration of reactant sulfuryl chloride as function of time t:  

ln[SO₂Cl₂] = ln[SO₂Cl₂]₀ - k∙t  

([SO₂Cl₂]₀ i the initial concentration and k is the rate constant)  

=>  

[SO₂Cl₂] = e^(ln[SO₂Cl₂]₀ - k∙t) = e^(ln[SO₂Cl₂]₀) ∙ e^( - k∙t)  

= [SO₂Cl₂]₀ ∙e^( - k∙t)  

The decomposition of SO₂Cl₂ follows the reaction equation  

SO₂Cl₂ → SO₂ + Cl₂  

So per mole of sulfuryl chloride disappeared one mole of sulfur dioxide is formed. Hence the changes in concentrations of these compounds are related as:  

-∆[SO₂Cl₂] = ∆{SO₂]  

<=>  

[SO₂Cl₂]₀ - [SO₂Cl₂] = [SO₂] - [SO₂]₀  

Assuming there was initially no sulfur dioxide you get for its concentration:  

[SO₂] = [SO₂Cl₂]₀ - [SO₂Cl₂]  

= [SO₂Cl₂]₀ - [SO₂Cl₂]₀ ∙e^( - k∙t)  

= [SO₂Cl₂]₀ ∙(1 - e^( - k∙t))  

after 210s starting with 0.150M of [SO₂Cl₂]  

[SO₂]= 0.150M ∙(1 - e^( - 1.48×10⁻⁴s⁻¹k ∙ 240s) =