Question

It takes 5.2 minutes for a 4.0 g sample of francium-210 to decay until only 1.0 g is left. What is the half-life of francium-210? 7.8 minutes 1.3 minutes 2.6 minutes 5.2 minutes

Answer 1

This problem is not about chemistry it is about math. the half-life of a certain thing is the time that it needs to decay 50% of what it had before. so if there is 4g, after 1 half there will be only 2g. so to become 1g it is passed 2 half-lives. 4,0g, 2,0g, 1,0g so it is the time divided by 2( numbers of half-lives). So the half-life of Francium 210 is 2,6 min.

Answer 2

Answer: The half life of the reaction is 2.6 minutes.

Explanation:

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,  

k = rate constant = ?

t = time taken for decay process = 5.2 minutes

$[A_o]$ = initial amount of the reactant = 4.0 g

[A] = amount left after decay process =  1.0 g

Putting values in above equation, we get:

$k=\frac{2.303}{5.2min}\log\frac{4.0}{1.0}\\\\k=0.267min^{-1}$

The equation used to calculate half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

We are given:

$k=0.267min^{-1}$

Putting values in above equation, we get:

$t_{1/2}=\frac{0.693}{0.267min^{-1}}=2.6min$

Hence, the half life of the reaction is 2.6 minutes.