Question

The half-life for the first order reaction,
N205 → 2N02 + 1/2 0, is 24 h at
30° C. Starting with 10 g of N2O5, how many
grams of N205 will remain after a period of
96 h?
(a) 1.25 g (b) 0.63 g (c) 1.77 g (d) 0.5 g​

Answer 1

Answer:

b) 0.63g

Explanation:

LONG METHOD

as half life is 24 hrs

so no. of half lives spent =96/24

now if we start with N grams at t=0

N➡️N/2➡️N/4➡️N/8➡️N/16

in four half lives hence amount left is N/16

here N=10

hence 10/16= 0.625 g

SHORT METHOD

After n half lives amount of substance left

= Initial amount/ 2^n

here n = 4

hence 10/16= 0.625

Hope I was able to solve your doubt