Question

Under certain conditions the rate of this reaction is zero order in ammonia with a rate constant of :

0.0067·Ms−1: 2N2O(g)→2N2(g)+O2(g)
Suppose a flask is charged under these conditions with of ammonia. After how much time is there only left? You may assume no other reaction is important. Be sure your answer has a unit symbol, if necessary, and round it to significant digits.

Answer 1

Step-by-step explanation:

From the given information:

`2NO_2_((g)) \to 2N_(2(g)) + O_(2(g))`

The above reaction is a zero-order reaction.

The rate constant = 0.0067 M-s⁻¹

Suppose the volume of the flask = 4L

Initial Mol of dinitrogen monoxide = 300 mmol

The final mol of dinitrogen monoxide = 150 mmol

The molarity of dinitrogen monoxide =
`\frac{ \text{number of moles of dinitrogen monoxide }}{\text{volume of flask}}`

`= (300 \ m)/( 4 L)`

= 0.0075 mmol/L

= 0.0075 L

The final concentration
`= (150 \ m)/( 4 L)`

= 0.0375 L

By applying zero order equation

`kt = [A_o] -[At]`

`(0.0067)(t) = 0.075 - 0.0375`

`(0.0067)(t) = 0.0375`

`\mathbf{t = 5.59 \ seconds}`