Question

The decomposition of phosphine,
`4PH_(3)(g) to P_(4)(g) + 6H_(2)(g)` has rate law , Rate = `k[PH_(3)]`. The rate constant is `6.0 xx 10^(-4) s^(-1)` at 300 K and activation energy is `3.05 xx 10^(5)` J `mol^(-1)`. Calculate the value of the rate constant at 310 K (`R= 8.314 JK^(-1)mol^(-1))`

Answer 1

According to Arrhenius equation.
`log k_(2)/k_(1) = E_(a)/(2.303R) [1/T_(1)-1/T_(2)]`
`E_(a) =3.05 xx 10^(5)J mol^(-1)`, `T_(1)=300K`, `T_(2)=300K`, `R=8.314 J mol^(-1)K^(-1)`
`logk_(2)/k_(1) = (3.05 xx 10^(5)J mol^(-1))/(2.303 xx 8.314 J mol^(-1))[(310-300)/(300 xx 310)]=1.7128`
`k_(2)/k_(1)`= Antilog 1.7128 = 51.6178
`k_(2)= 51.6178 xx k_(1) = 51.6178 xx k_(1) = 51.6178 xx (6.0 xx 10^(-4)s^(-1)) = 30.97 xx 10^(-3)s^(-1)`.