1.

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»

Solution :(i) Order of reaction =`1, t_(1//2)= 60, "seconds" , k = ? `
We know that , `k=(2.0303)/(t_(1//2))`
`k = (2.303)/60 = 0.01155s^(-1)`
(II) `[A_0]=100% , t = 180 s, k = 0.01155 " seconds "^(-1) , [A] = ? `
For the FIRST order reaction `k=(2.303)/t log.([A]_0)/([A])`
`0.01155=(2.303)/180log.(100/([A]))IMPLIES(0.1155xx180)/(2.303)=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%


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