Chemistry, asked by varneyamos35, 7 hours ago

The rate at which tree crickets chirp is 2.0 × 102 per minute at 27°C but only 39.6 per minute at 5°C. From these data, calculate the "activation energy" for the chirping process. (Hint: The ratio of rates is equal to the ratio of rate constants.)​

Answers

Answered by Rudramanojkumarghume
0

Answer:

square of 36 and please follow

Answered by shilpa85475
2

Given,

The rate at 27°C(r_{1})=2.0\times 10^{2} min^{-1}

The rate at 5°C(r_{2})=39.6 min^{-1}

The ratio of the rates is equal to the ratio of rate constant:

\frac{r_{1} }{r_{2} }= \frac{k_{1} }{k_{2} }

Solving for the activation energy of the chirping process:

ln\frac{k_{1} }{k_{2} } =\frac{-E_{a} }{R}(\frac{1}{T_{1} }-\frac{1}{T_{2} } )

E_{a}=\frac{-Rln(k_{1}/k_{2})  }{\frac{1}{T_{1} }-\frac{1}{T_{2} }   }

E_{a}=\frac{-8.3145J/mol.Kln\frac{2\times 10^{2}  }{39.6}   }{\frac{1}{(27+273) }-\frac{1}{(5+273) }   }=51045J

Hence, the "activation energy" for the chirping process is 51 kJ/mol.

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