Question

In a particular experiment at 300 °C, [NO₂] drops from 0.0295 to 0.00769 M in 316 s . The rate of disappearance of NO₂ for this period is ________ M/s.

A) 1.38×10⁻⁴
B) −1.18×10⁻⁴
C) 3.45×10⁻⁵
D) 6.90×10⁻⁵
E) 1.45×10⁴

Answer 1

The rate of disappearance of NO₂ for the given time period is 6.90 x 10⁻⁵ M/s.

Step-by-step explanation:

The rate of disappearance of NO₂ can be determined using the formula:

Rate = (change in concentration of NO₂) / (change in time)

Using the given data, the initial concentration of NO₂ is 0.0295 M and the final concentration is 0.00769 M. The time interval is 316 seconds.

Plugging in these values into the formula, we get:

Rate = (0.0295 - 0.00769) / 316

Rate = 0.02181 / 316

Rate = 6.90 x 10-5 M/s