In a certain gaseous reaction A`rarr` B, the initial pressure is 214 atm and the rate constant is `2.303 xx 10^(- 4) s^(-1)` . What would be pressure (in atm) of A after 5 mins? [Given:`10^(0.03) ` = 1. 07]

Answer

200

Explanation

Initial pressure `P_(i) `= 214 atm
Rate constant `k = 2. 303 xx 10^(- 4) s^(-1)`
Unit of rate constant indicates first order reaction.
Time, t = 5minutes = `5 xx 60`s
The integrated rate law is,
`k = (2.303)/(t ) log_(10) ((P_(i))/(P_(f)))`
` 2.303 xx 10^(-4) = (2.303)/(5xx 60) "log"_(10) ((214)/(P_(f)))`
`:. 0.03 = "log"_(10) ((214)/(P_(r )))`
`:. P_(f) = (214)/(1.07) = 200 ` atm