how do i solve the following problem?
Nitrogen dioxide decomposes to nitric oxide and oxygen via the reaction:
2NO2 → 2NO + O2
In a particular experiment at 300°C, [NO2] drops from 0.0100 to 0.00650 M in 100 s. The rate of
Disappearance of NO2 for this period is __________ M/s.
3.5x10^-5 is supposedly the answer, but how do i get to that answer?
(0.0100-0.00650)/100 =
(.1M-.0065M)/(100sec)=3.5e-5
To find the rate of disappearance of NO2, you need to calculate the change in concentration of NO2 per unit of time. This can be done using the formula:
Rate = (change in concentration of NO2) / (change in time)
Given:
Initial concentration of NO2, [NO2]initial = 0.0100 M
Final concentration of NO2, [NO2]final = 0.00650 M
Change in time, Δt = 100 s
Now, we can substitute these values into the formula:
Rate = ([NO2]final - [NO2]initial) / Δt
Rate = (0.00650 M - 0.0100 M) / 100 s
Simplifying the equation:
Rate = -0.0035 M / 100 s
Rate = -3.5x10^-5 M/s (Note the negative sign indicates the rate of disappearance)
Therefore, the rate of disappearance of NO2 for this period is 3.5x10^-5 M/s.
To find the rate of disappearance of NO2, you can use the formula:
Rate = ∆[NO2] / ∆t
Where ∆[NO2] is the change in concentration of NO2 and ∆t is the change in time.
Given:
Initial concentration, [NO2]initial = 0.0100 M
Final concentration, [NO2]final = 0.00650 M
Time interval, ∆t = 100 s
First, find the change in concentration (∆[NO2]):
∆[NO2] = [NO2]final - [NO2]initial
= 0.00650 M - 0.0100 M
= -0.0035 M
Note the negative sign, indicating a decrease in concentration.
Next, substitute these values into the rate formula:
Rate = ∆[NO2] / ∆t
= -0.0035 M / 100 s
= -3.5 × 10^-5 M/s
It seems there was an error in the answer you provided. The correct rate of disappearance of NO2 for this period is -3.5 × 10^-5 M/s.