Question

The decomposition of N2O5 in solution in carbon tetrachloride proceeds via the reaction 2 N2O5(soln) → 4 NO2(soln) + O2(soln) The reaction is first order and has a rate constant of 4.82 × 10-3 s-1 at 64°C. If the reaction is initiated with 0.085 mol in a 1.00-L vessel, how many moles remain after 151 s?

Answer 1

Answer: The amount remained after 151 seconds are 0.041 moles

Explanation:

All the radioactive reactions follows first order kinetics.

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  = $4.82\times 10^{-3}s^{-1}$

t = time taken for decay process = 151 sec

$[A_o]$ = initial amount of the reactant = 0.085 moles

[A] = amount left after decay process =  ?

Putting values in above equation, we get:

$4.82\times 10^{-3}=\frac{2.303}{151}\log\frac{0.085}{[A]}$

$[A]=0.041moles$

Hence, the amount remained after 151 seconds are 0.041 moles