Question

The reaction NO₂(g) + NO₂(g) →N₂O₄ (g). the rate constant for this reaction at a certain temperature is 0.025 m⁻¹s⁻¹. if the initial concentration of NO₂ is 0.72 m, how many seconds will it take for its concentration to decrease to 0.31 m?

Answer 1

To determine how long it will take for the concentration of NO₂ to decrease from 0.72 M to 0.31 M, we apply the second-order integrated rate law. By plugging the initial concentration, final concentration, and rate constant into the formula, we calculate that it will take approximately 73.48 seconds.

Step-by-step explanation:

The question asks for the time required for the concentration of NO₂ to decrease from 0.72 M to 0.31 M in the reaction NO₂(g) + NO₂(g) → N₂O₄ (g). Since the reaction is second-order with respect to NO₂, we can use the second-order integrated rate law: 1/[NO₂] - 1/[NO₂]0 = kt. We have the initial concentration ([NO₂]0 = 0.72 M), the final concentration ([NO₂] = 0.31 M), and the rate constant (k = 0.025 M⁻¹s⁻¹). Substituting these values into the equation, we can solve for the time (t).

Calculating the time (t):

1. 1/[NO₂] - 1/[NO₂]0 = kt
2. 1/0.31 M - 1/0.72 M = (0.025 M⁻¹s⁻¹)t
3. t = (1/0.31 - 1/0.72) / 0.025
4. t = (3.226 - 1.389) / 0.025
5. t = 1.837 / 0.025
6. t = 73.48 s

Therefore, it will take approximately 73.48 seconds for the concentration of NO₂ to decrease from 0.72 M to 0.31 M at the given temperature and rate constant.