The half life for the first order reaction N₂O₅ → 2NO₂ + 1/2O₂ is 24 hrs. at 30°C. Starting with 10g of N₂O₅ how many grams of N₂O₅ will remain after a period of 96 hours?
(1) 1.25 g                  (2) 0.63 g               (3) 1.77 g              (4) 0.5 g

Dear student,
For a 1^st order reaction,
t_1/2= 0.693/k  (k=rate constant)
Therefore, k = 0.693/t_1/2 = 0.693/24= 0.0289 hr^-1
kt= 2.303*log(a/(a-x))  {a= initial amount, a-x= amount left}
0.0289*96 = 2.303*log(10/(10-x))
log(10/(10-x))= 1.2
10/(10-x) = antilog1.2
10/(10-x) = 15.85 
10 = 158.5-15.85x
15.85x = 148.5
x= 9.36g
a-x = 10-9.36= 0.63g
Hence, option 2 is correct
Regards