Using the Arrhenius equation : The rate constant fot the formation of hydrogen iodide from the elemewnts
`H_(2)(g)+I_(2)(g)rarr2HI(g)`
is `2.7xx10^(-4)L//(mol.s)at 600 k and 3.5xx10^(-3)L//(mol.s) at 650 k` .
(a) Find the activation energy `E_(a)` . (b) Calculate the rate constant at 700 k .
Strategy : (a) Substitute the data given in the problem statement into the Equation `(4.37)` noted just before this example, then solve for `E_(a)` . , (b) Use the same equation, but substitute for `k_(1), T_(1), T_(2)` and `E_(a)` obtained in (a) and solve for `k_(2)` .

Explanation

According to equation `(4.37)`
`log ((k_(2))/(k_(1))) = ((-E_(a))/(2.303 R)) ((1)/(T_(2)) - (1)/(T_(1)))`
or `log ((k_(2))/(k_(1))) = ((E_(a))/(2.303 R)) ((1)/(T_(1)) - (1)/(T_(2)))`
Substituting the data gives
`log ((3.5xx10^(-3))/(2.7xx10^(-4))) = (E_(a))/(2.303xx8.314 J//(mol.k)) ((1)/(600k) - (1)/(650 k))`
`log (1.30xx10^(-1)) = 1.11= (E_(a))/(2.303xx8.314 J mol^(-1)) xx (1.28xx10^(-4))`
Hence, `E_(a) = ((1.11)(2.303)(8.314 J mol^(-1)))/(1.28xx10^(-4))`
`= 16.6xx10^(4)J mol^(-1)`
`= 1.66xx10^(5) J mol^(-1)`
(b) Substitute
`E_(a) = 1.66xx10^(5)J mol^(-1)`
`k_(1) = 2.7xx10^(-5) L//(mol.s)`
`k_(2) = "unkonown as yet" , T_(1) = 60 k`
we get
`log (k_(2))/(2.7xx10^(-4)L//(mol.s)) = (1.66xx10^(5)J mol^(-1))/(2.303xx8.314 J mol^(-1) k^(-1)) ((1)/(600 k) - (1)/(700 k))`
`=2.07`
Taking antilogarithm
`((k_(2))/(2.7xx10^(-4)L//(mol.s)))= 10^(2.07) = 1.2xx10^(2)`
Hence, `k_(2) = (1.2xx10^(2))(2.7xx10^(-4))L//mol.s`
`=3.2xx10^(-2)L//(mol.s)`