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Consider a first order gas phase decompo...

Consider a first order gas phase decomposition reaction given below: `A_((g)) to B_((g)) +C_((g))`
The initial pressure is 6.0 atm. The pressure drops to 3.0 atm after 6.93 min. How much time (in minutes) would it take to lower the partial pressure of `A_((g))` by 4.0 atm? [Consider: `log_(10) (3) = 0.48`]

Text Solution

Verified by Experts

The correct Answer is:
`11.05`
Since the partial pressure reduces to half after 6.93 minutes, the half-life, `t_(1//2)` = 6.93 min
`k=(0.693)/(t_(1//2))=(0.693)/(6.93)=0.10"min"^(-1)`
For a gas phase first order reaction,
`k=(2.303)/(t) log_(10) ((p_(i))/(p_(A)))`
`p_(i)=6.0"atm", p_(A) =6.0-4.0=2.0` atm
`:.0.10=(2.303)/(t)log_(10)((6.0)/(2.0))`
`:.t=23.03log(3.0) =11.05` min
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