Time required to decompose $S O_{2} C L_{2}$ to half of its
initial amount is 60 minutes . If the decompose is a first order reaction,
calculate the rate constant of the
reaction.

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Solution

Given $∴ t_{1 / 2} = 60 m i n = 60 \times 60 = 3600 s e c$
$∵$ we know that 1 min = 60 sec
Half life for the $1^{s t}$ order reaction,
$t_{1 / 2} = \frac{0.693}{k}$
$3600 = \frac{0.693}{k}$
$k = \frac{0.693}{3600} = 0.0001925 = 1.925 \times 10^{- 4} s^{- 1}$

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Time required to decompose SO2CL2 to half of itsinitial amount is 60 minutes . If the decompose is a first order reaction,calculate the rate constant of thereaction.

Given ∴t1/2=60min=60×60=3600sec∵ we know that 1 min = 60 secHalf life for the 1st order reaction,t1/2=0.693k3600=0.693kk=0.6933600=0.0001925=1.925×10−4s−1

Time required to decompose $S O_{2} C L_{2}$ to half of its
initial amount is 60 minutes . If the decompose is a first order reaction,
calculate the rate constant of the
reaction.

Given $∴ t_{1 / 2} = 60 m i n = 60 \times 60 = 3600 s e c$
$∵$ we know that 1 min = 60 sec
Half life for the $1^{s t}$ order reaction,
$t_{1 / 2} = \frac{0.693}{k}$
$3600 = \frac{0.693}{k}$
$k = \frac{0.693}{3600} = 0.0001925 = 1.925 \times 10^{- 4} s^{- 1}$