Question

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

Answer 1

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:

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,  

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

Thus,

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

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

246.67 min is the half-life of the reaction.

Answer 2

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