The rate of popcorn popping at different temperatures was found to be described by the Arrhenius equation. How many times faster does the popcorn pop at 210°C compared with that at 180°C when the activation energy is 167 kJ/mol?

Question

Quizplus · Accepted Answer

The rate of reaction according to the Arrhenius equation is given by $k = Ae^{-E_a/(RT)}$, where $k$ is the rate constant, $A$ is the pre-exponential factor, $E_a$ is the activation energy, $R$ is the gas constant (8.314 J/mol·K), and $T$ is the temperature in Kelvin. To find how many times faster the reaction occurs at one temperature compared to another, we calculate the ratio of the rate constants at these temperatures. Given: $E_a = 167 \times 10^3 \, \text{J/mol}$, $T_1 = 210°C = 483.15K$, and $T_2 = 180°C = 453.15K$.The ratio of the rates is given by $\frac{k_2}{k_1} = \frac{e^{-E_a/(RT_2)}}{e^{-E_a/(RT_1)}} = e^{E_a/R(1/T_1 - 1/T_2)}$.Plugging in the values: $e^{(167000)/(8.314)(1/453.15 - 1/483.15)} \approx 15.7$.Therefore, the popcorn pops approximately 15.7 times faster at 210°C compared to 180°C.

The Rate of Popcorn Popping at Different Temperatures Was Found ×\times×

The Rate of Popcorn Popping at Different Temperatures Was Found $\times$