Question

Socl2 decomposes to half of its initial amount in 60 min.Calculte the rate constant

Answer 1

Answer:

We know that t(half ) for the first order reaction is \frac{0.693}{k} = t_{1/2}

and we have given the value of half time t_{1/2} = 60 min

thus, k = \frac{0.693}{60}min^{-1}

= 0.01155 /min

OR = 1.1925\times 10^{-4}sec

Alternative method

we can also solve this problem by using the first order reaction equation.

k = \frac{2.303}{t}\log\frac{[R]_{0}}{[R]}

put [R]=[R_{0}]/2

Answer 2

Answer:

... to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the rea