Question

For the reaction: 2N₂O₅(g) → ₄NO₂(g) + O₂(g), the rate law is: rate = k[N₂O₅]. At 300 K, the half-life is 2.50 × 10⁴ seconds and the activation energy is 103.3 kJ/mol. What is the half-life at 310 K?

Answer 1

To find the half-life at 310 K for the given reaction, we can use the Arrhenius equation to calculate the rate constant at the new temperature. Then, we can use the rate constant to determine the new half-life using the equation: t₂ = ln(2)/k₂.

Step-by-step explanation:

The rate law for the reaction is given as: rate = k[N₂O₅]. In order to find the half-life at 310 K, we can use the Arrhenius equation: k₂/k₁ = e^((Ea/R)*(1/T₁ - 1/T₂)), where k₁ and T₁ are the rate constant and temperature respectively at the given conditions, k₂ and T₂ are the rate constant and temperature respectively at the new conditions, Ea is the activation energy, and R is the gas constant. Since we are given the activation energy, rate constant, and temperature at one condition, we can plug in the values to calculate the rate constant at the new condition. Then, we can use the rate constant to find the new half-life using the equation: t₂ = ln(2)/k₂.