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The decomposition of phosphine, 4PH(3) (...

The decomposition of phosphine, `4PH_(3) (g) to P_(4) (g) +6H_(2)(g)` has the 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 rate constant at 310 K. `("Given, R=8.314 JK"^(-1)" mol"^(-1))`

Text Solution

Verified by Experts

Given, `T_(1)=300K, T_(2)=310K`
`E_(alpha)=3.05 xx 10^(5)" J mol"^(-1)`
`k_(1)=6.0 xx 10^(-4) s^(-1), k_(2)=?`
According to Arrhenius equation,
`log [k_2/k_1]=(E_(alpha))/(2.303 R) [(T_2-T_1)/(T_1T_2)]`
`log [k_2/k_1]=(3.05 xx 10^(5))/(2.303 xx 8.314) [(310-300)/(300 xx 310)]`
`log [k_2/k_1] =(3.05 xx 10^(5) xx 10)/(2.303 xx 8.314 xx 300 xx 310)`
`k_(2)/k_(1)="antilog 1.7128"`
`k_(2)/k_(1)=5.162 xx 10^(1) =51.62`
`k_(2)/(6.0 xx 10^(-4))=5162 or k_2=3.0972 xx 10^(-2) s^(-1)`
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