The chirping of tree crickets has sometimes been used to predict temperature. At

Question

The chirping of tree crickets has sometimes been used to predict temperature. At 25.0 degree C the rate of chirping was recorded to be 179 chirps per minute. At 21 9 degree C it was recorded to be 142 chirps/min. Calculate the activation energy for this reaction. If a temperature increase from 10.0 degree C to 20.0 degree C doubles the rate constant for a reaction, what is the value of the activation energy for the reaction? Increasing the temperature of a reaction from 20.0 degree C to an unknown temperature triples the rate constant for a reaction. Given that the activation energy for the reaction is 54.95 kJ/mol, determine the unknown temperature in degree C.

Explanation / Answer

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2)
T1 = 25 oC = (25+273) k = 298 K
T2 = 22.9 oC = (273 + 22.9) = 295.9 K
k1 = 179
k2 = 142
ln(K2/K1) = Ea/R * (1/T1 - 1/T2)
ln(142/179) =Ea/8.314 * (1/298 - 1/295.9)
-0.2316 = Ea/8.314 * (-2.382*10^-5)
Ea = 80852 J/mol
= 80.852 KJ/mol

Answer: 80.852 KJ/mol

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