Question

A first order reaction is 20% complete in 5 mins. calculate the time for the reaction to be 60% complete.

Answer 1

t = 2.303/k log(Ninitial/N final)

5= 2.303/k log(100/80)

By solving log.....

K =2.303/5(0.1)........ (I)

Now,.. According to the question...

t=2.303/k log (100/40)

Putting value of k from eq. 1

t=50(0.4)

t= 20min

Answer 2

The time for the reaction to be 60% complete is 20 minutes.

Explanation:

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant  = ?

t = time for decay  = 5 minutes

a = let initial amount of the reactant  = 100 g

x = amount decayed = 20 g

a - x = amount left after decay process  = 100 - 20 = 80

Now put all the given values in above equation, we get

$5=\frac{2.303}{k}\log\frac{100}{80}$

$k=0.045min^{-1}$

Now time for the reaction to be 60% complete:

$t=\frac{2.303}{0.045}\log\frac{100}{100-60}$

$t=20minutes$

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