Question

In the reaction of gaseous N₂O₅ to yield NO₂ gas and O₂ gas as shown below the following data table is obtained:

2 N₂O₅ (g) → 4 NO₂ (g) + O₂ (g)

	Data Table #1
Time (sec)	[N2O5]	[O2]
0	0.200 M	0
300	0.182 M	0.009 M
600	0.166 M	0.017 M
900	0.152 M	0.024 M
1200	0.140 M	0.030 M
1800	0.122 M	0.039 M
2400	0.112 M	0.044 M
3000	0.108 M	0.046 M

Complete the following three problems:

Using the [O₂] data from the table show the calculation of the average rate over the measured time interval from 0 to 3000 secs.

Using the [O₂] data from the table show the calculation of the instantaneous rate late in the reaction (2400 secs to 3000 secs).

Explain the relative values of the average rate and the late instantaneous rate.

Answer 1

We calculate the average rate of a reaction over a time interval by dividing the change in concentration over that time period by the time interval. For the change in concentration of a reactant, the equation, where the brackets mean "concentration of", is

rate = [0.046 M] - [0] / 3000 s - 0

average rate = 1.5 x10^-5 M / s

An instantaneous rate is the rate at some instant in time.

We determine an instantaneous rate at time t:

• by calculating the negative of the slope of the curve of concentration of a reactant versus time at time t.
• by calculating the slope of the curve of concentration of a product versus time at time t.

the slope would be =instantaneous rate = [O₂]₃₀₀₀ - [O₂]₂₄₀₀ / t₃₀₀₀ - t₂₄₀₀

instantaneous rate = 0.046 M - 0.044 M / 3000 s - 2400 s

instantaneous rate = 3.3 x10^-6 M/s

The average rate give us the variation of the concentration during the complete transformation, That will take into account all the reaction so it is logical that is bigger than the instantaneous rate that will give us the information of only one small fraction of the reaction.

When we plote the M vs t we will not obtain a linear plot, so the velocity along the line will be different, and normally at the end of the time it will be a slower reaction.