InterviewSolution
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InterviewSolution
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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)` . |
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Answer» 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)` |
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