For a certain reaction, the variation of the rate constant with temperature is given as: ln K t = ln K 0 + 0.11 t (for t 0℃ ) where K t and K 0 are the rate constants at t ℃ and 0℃ , respect

Question

For a certain reaction, the variation of the rate constant with temperature is given as:

$ln K_{t} = ln K_{0} + 0.11 t$ (for $t > 0℃$ )

where $K_{t}$ and $K_{0}$ are the rate constants at $t ℃$ and $0℃$ , respectively. The value of the temperature coefficient of the reaction, assuming it is constant in the temperature range of interest, is

(given $ln 2 = 0.693$ , $ln 3 = 1.1$ ).

TestHub · Accepted Answer

The temperature coefficient is K t + 10 ​ / K t ​ . From ln K t ​ = ln K 0 ​ + 0.11 t , we get K t ​ = K 0 ​ e 0.11 t . So, K t + 10 ​ / K t ​ = ( K 0 ​ e 0.11 ( t + 10 ) ) / ( K 0 ​ e 0.11 t ) = e 0.11 × 10 = e 1.1 . Given ln 3 = 1.1 , so e 1.1 = 3 .

Chemistry - Chemical Kinetics Question with Solution | TestHub

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