In a certain gaseous reaction A longrightarrow B, the initial pressure is 214 atm and the rate constant is 2.303 × 10-4 s -1. What would be pressure (in atm ) of A after 5 mins? [Given: 100.03=1.07 ]

Question

In a certain gaseous reaction A longrightarrow B, the initial pressure is 214 atm and the rate constant is 2.303 × 10-4 s -1. What would be pressure (in atm ) of A after 5 mins? [Given: 100.03=1.07 ] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Initial pressure Pi=214 atm Rate constant k =2.303 × 10-4 s -1 Unit of rate constant indicates first order reaction. Time, t =5 minutes =5 × 60 s The integrated rate law is, k =(2.303/ t ) log 10(( P i / P f )) 2.303 × 10-4=(2.303/5 × 60) log 10((214/ P f )) ∴ 0.03= log 10((214/ P f )) ∴ 1.07=(214/ P f ) ∴ P f =(214/1.07)=200 atm

Q. In a certain gaseous reaction A⟶B, the initial pressure is 214atm and the rate constant is 2.303×10−4s−1. What would be pressure (in atm) of A after 5 mins? [Given: 100.03=1.07 ]

Q. In a certain gaseous reaction $A ⟶ B$ , the initial pressure is $214 a t m$ and the rate constant is $2.303 \times 1 0^{- 4} s^{- 1}$ . What would be pressure (in $a t m)$ of $A$ after $5$ mins?
[Given: $1 0^{0.03} = 1.07$ ]