Question

The time for half change in a first order decomposition of a substance A is 60 seconds. Calculate the rate constant. How much of A will be left after 180 seconds?​

Answer 1

Answer:

Order of reaction =1,t1/2=60,seconds,k=?

We know that , k=2.0303t1/2

k=2.30360=0.01155s−1

(ii) [A0]=100%,t=180s,k=0.01155 seconds −1,[A]=?

For the first order reaction k=2.303tlog.[A]0[A]

0.01155=2.303180log.(100[A])⇒0.1155×1802.303=log(100[A])

0.9207=log100-log[A]

log [A] = log 100 - 0.9207

log [A] = 2 - 0.9207

log [A] =1.0973

[A] = antilog of (1.0973)

[A] = 12.5%

Answer 2

Explanation:

1. Order of a reaction = 1 t1/2 = 60 seconds, k = ? k = 2.303 60 2.30360 We know that, k = 2.303 t 1 / 2 2.303t1/2 k = 2.303 60 2.30360 = 0.01155 s-1 2. [A0] = 100% t = 180 s k = 0.01155 seconds-1 [A] = ? For the first order reaction k = 0.9207 = log 100 – log [A] log [A] = log 100 – 0.9207 log [A] = 2 – 0.9207 log[A] = 1.0973 [A] = antilog of (1.0973) [A] = 12.5