Questions: Under certain conditions the rate of this reaction is zero order in dinitrogen monoxide with a rate constant of 0.0059 M · s^-1: 2 N2O(g) → 2 N2(g) + O2(g) Suppose a 3.0 L flask is charged under these conditions with 400 mmol of dinitrogen monoxide. After how much time is there only 200 mmol left? You may assume no other reaction is important. Be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.

Solution Steps

For a zero-order reaction, the rate of reaction is constant and independent of the concentration of the reactant. The rate law for a zero-order reaction is given by:

\[ \text{Rate} = k \]

where \( k \) is the rate constant. The integrated rate law for a zero-order reaction is:

\[ [A] = [A]_0 - kt \]

where \([A]\) is the concentration of the reactant at time \( t \), \([A]_0\) is the initial concentration, and \( k \) is the rate constant.

The initial amount of dinitrogen monoxide is 400 mmol, and the final amount is 200 mmol. We need to convert these amounts to concentrations in molarity (M).

\[ [A]_0 = \frac{400 \, \text{mmol}}{3.0 \, \text{L}} = \frac{0.400 \, \text{mol}}{3.0 \, \text{L}} = 0.1333 \, \text{M} \]

\[ [A] = \frac{200 \, \text{mmol}}{3.0 \, \text{L}} = \frac{0.200 \, \text{mol}}{3.0 \, \text{L}} = 0.0667 \, \text{M} \]

Substitute the known values into the integrated rate law to solve for \( t \):

\[ 0.0667 \, \text{M} = 0.1333 \, \text{M} - (0.0059 \, \text{M} \cdot \text{s}^{-1}) \cdot t \]

Rearrange to solve for \( t \):

\[ t = \frac{0.1333 \, \text{M} - 0.0667 \, \text{M}}{0.0059 \, \text{M} \cdot \text{s}^{-1}} \]

\[ t = \frac{0.0666 \, \text{M}}{0.0059 \, \text{M} \cdot \text{s}^{-1}} \approx 11288.1 \, \text{s} \]

Final Answer