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The half life for the first order reaction N_{2}O_{5} → 2NO_{2} + 1/2O_{2} is 24 hrs. at 30°C. Starting with 10g of N_{2}O_{5} how many grams of N_{2}O_{5} will remain after a period of 96 hours?

(1) 1.25 g (2) 0.63 g (3) 1.77 g (4) 0.5 g

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

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