The flashing of fireflies results from a chemical reaction, and the Arrhenius equation can describe the rate of flashing. A batch of fireflies was observed to flash at a rate of 17.0 times per minute at 25°C and 5.0 times per minute at 15°C. Use these data to calculate the apparent activation energy for the reaction causing the flashing.

To calculate the apparent activation energy (Ea) using the Arrhenius equation, we need to understand the equation itself first. The Arrhenius equation is as follows:

k = Ae^(-Ea/RT)

where: - k is the rate constant, - A is the frequency factor, - Ea is the activation energy, - R is the gas constant (8.314 J/mol·K), and - T is the temperature in Kelvin.

The equation that relates two rate constants at two different temperatures (T1 and T2) is:

ln(k2/k1) = (-Ea/R) * (1/T2 - 1/T1)

Given: k1 = 17.0 min^-1 at T1 = 25°C k2 = 5.0 min^-1 at T2 = 15°C

First, we need to convert temperatures from Celsius to Kelvin: T1 = 25°C = 25 + 273.15 = 298.15 K T2 = 15°C = 15 + 273.15 = 288.15 K

Now, we can plug these values into the modified Arrhenius equation:

ln(k2/k1) = (-Ea/R) * (1/T2 - 1/T1)

ln(5.0/17.0) = (-Ea/8.314) * (1/288.15 - 1/298.15)

Calculating the left side of the equation:

ln(5.0/17.0) = ln(0.2941) = -1.2238

Now solve for Ea:

-1.2238 = (-Ea/8.314) * (1/288.15 - 1/298.15)

-1.2238 = (-Ea/8.314) * (0.003472 - 0.003355)

-1.2238 = (-Ea/8.314) * (0.000117)

Now, solve for Ea by isolating it on one side:

Ea = -1.2238 * 8.314 / 0.000117

Ea = -103.610172 / 0.000117

Ea ≈ 88581.8 J/mol

Ea ≈ 88.6 kJ/mol (since 1 J = 1 x 10^-3 kJ)

Therefore, the apparent activation energy for the reaction causing the flashing in fireflies is approximately 88.6 kJ/mol