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SO2Cl2 decomposes by first order kinetics and k=2,81x106-3 min-1 at a given temperature. The initial concentration of SO2Cl2=0.015M. Determine the half-life of the reaction

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Answer:

246.67 min

Explanation:

Half life is the time at which the concentration of the reactant reduced to half.

Half life expression for first order kinetic is:

[tex]t_{1/2}=\frac {ln\ 2}{k}[/tex]

Where, k is rate constant

So,  

Given that;- k = [tex]2.81\times 10^{-3}\ min^{-1}[/tex] (Values corrected from source)

Thus,

[tex]t_{1/2}=\frac{\ln2}{2.81\times 10^{-3}\ min^{-1}}=\frac{\ln2}{2.81\times 10^{-3}}\ min[/tex]

[tex]t_{1/2}==355.87188\ln \left(2\right)\ min=246.67\ min[/tex]

246.67 min is the half-life of the reaction.

Answer:

t(1/2) = 246.67 min

Explanation:

a: SO2Cl2

⇒ - ra = k(Ca)∧α = - δCa/δt

∴ k = 2.81 E-3 min-1

∴ α = 1

⇒ Ln(Cao/Ca) = k*t

half-life of the reaction:

∴ Ca = (1/2)Cao

⇒ Ln (2) = (2.81 E-3 min-1)*t

⇒ t (1/2) = ( 0.6931 ) / ( 2.81 E-3 min-1 )

⇒ t (1/2) = 246.67 min

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