Question

The half life for the first order reaction $N_2{O}_5\to 2N{O}_2+\frac{1}{2}{O}_2$ is $24$ hrs. at ${30}^{\circ }C$. Starting with $10g$ of $N_2{O}_5$ how many grams of $N_2{O}_5$ will remain after a period of $96$ hours?

• A. $1.25g$
• B. $0.63g$
• C. $1.77g$
• D. $0.5g$

Answer 1

Answer: (B)

$0.63g$

For a 1st order reaction,
t1/2= 0.693/k (k=rate constant)
Therefore, k = 0.693/t1/2 = 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