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The rate constant for the decomposition ...

The rate constant for the decomposition of hydrocarbons is `2.418 xx 10^(-5) "s"^(-1)` at 546 K. If the energy of activation is `179.9" kJ mol"^(-1)`, what will be the value of pre-exponential factor?

Text Solution

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According to Arrhenius equation,
`log k = log A - (E_a)/( 2.303 RT)`
We have, `k = 2.418 xx 10^(-5)" s"^(-1)`
`E_a = 179.9" kJ mol"^(-1)` or `179900" J mol"^(-1)`
`R = 8.314" JK"^(-1)" mol"^(-1), T = 546" K"`
`rArr log A = log k + (E_a)/( 2.303 RT)`
`= log (2.418 xx 10^(-5)" s"^(-1) ) + (179900" J mol"^(-1) )/( 2.303 xx (8.314" JK"^(-1)" mol"^(-1) ) xx 546" K"`
`log A = - 4.6184 + 17.21 = 12.5916`
` A =" antilog" 12.5916 = 3.9 xx 10^(-12)" s"^(-1)`
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