Question

What is the half-life of A? What will the pressure, initially 32.1 kPa, at

(a) 10 s,
(b) 10 min after initiation of the reaction?

Answer 1

a) 32.09 kPa

b) 32.09 kPa

Step-by-step explanation:

Given data:

rate constant
`= 3.56* 10^(-7) s^(-1)`

initial pressure is = 32.1 kPa

half life of A is calculated as

`t_(1/2) = (ln 2)/(k)`

`t_(1/2) = (ln 2)/(3.56\time 10^(-7))`

`t_(1/2) = = 1.956 * 10^6 s`

for calculating pressure we have follwing expression

`ln p = ln P_o - kt`

`P =P_o e^(-kt)`

a)
`P = 32.1 e^{-3.56* 10^(-7) * 10} = 32.09 kPa`

b)
`P = 32.1 e^{-3.56* 10^(-7) * 10* 60} = 32.09 kPa`