Studying the Rate of Reaction: O(g)+NO2(g) --> NO(g)+O2(g)

[viciousp]

Homework Statement

The rate of teaction
O(g) + NO2(g) --> NO(g) +O2(g)
was studied at a certain temperature.
NO2 is in a large excess at 1.0x10^13 molecules/cm^3

Data:
Time......[O](atoms/cm^3)
0.......5.0x10^9
1.0X10^-2......1.9X10^9
2.0X10^-2.....6.8X10^8
3.0X10^2......2.5^10^8

The reaction rate is know to be first order with respect to NO2. Determine the overall rate law and value of rate law constant

Homework Equations

k[A]=rate

The Attempt at a Solution

Ok I know that a O is a first order rate so the overall rate law is k[O][NO2]=Rate. The slope of ln[O] vs t is 99.8. So now how do you find the constant?

I think its k(1.0x10^13)(5.0x10^9)=99.8
so k= 2.0x10^-21 cm^3/atoms * s ,but something doesn't seem right about it

 

[nate808]

.

Dear fellow scientist,

Thank you for sharing your attempt at solving this problem. Your initial approach is correct - the overall rate law for this reaction is indeed first order with respect to both O and NO2. However, your calculation for the rate law constant is not quite correct.

To determine the value of the rate law constant, we can rearrange the rate law equation k[O][NO2]=rate to solve for k. In this case, since the reaction is first order with respect to both O and NO2, the rate law can be simplified to k[NO2]=rate. Now, we can use the data provided to plug in values for [NO2] and the corresponding reaction rates to solve for k.

Using the first data point at time 0, we have [NO2]=1.0x10^13 molecules/cm^3 and [O]=5.0x10^9 atoms/cm^3. Plugging these values into the rate law equation, we get: k(1.0x10^13)=5.0x10^9. Solving for k, we get k=5.0x10^-4 cm^3/molecule*s.

We can repeat this process for the other data points and we should get a consistent value for k. This is because the rate law is independent of the initial concentrations of the reactants, and only depends on the rate at which the reactants are consumed over time.

I hope this helps you in your analysis of this reaction. Keep up the good work!

 

[Blueshift5]

.

Thank you for sharing your work and thought process. It seems like you have the right idea, but there are a few steps that could use some clarification.

Firstly, when determining the overall rate law for a reaction, it is important to consider the rate-determining step. In this case, the given data suggests that the reaction is first order with respect to both O and NO2, but we cannot assume that this is true for the overall reaction without further information. It is possible that the overall reaction may be controlled by a different step that involves a different rate order for O and NO2.

Secondly, the units of the rate constant should be consistent with the units of the overall rate law. In this case, the units of the rate constant should be cm^3/molecule * s, since the units of the overall rate law are molecules/cm^3 * s. This can be achieved by dividing both sides of the equation by the concentration of NO2, since it is in excess and therefore its concentration does not change significantly over time.

To calculate the rate constant, you can use the slope of the ln[O] vs t graph. However, you will need to convert the slope from units of ln[O] to units of [O]. This can be done by taking the antilog of the slope, since ln(x) and e^x are inverse functions. The antilog of 99.8 is approximately 4.0x10^43. So, your final equation for the rate constant should be k = (4.0x10^43 cm^3/molecule * s)/(1.0x10^13 molecules/cm^3) = 4.0x10^30 cm^3/molecule * s.

Overall, your approach was on the right track, but it is important to consider the rate-determining step and ensure that the units are consistent throughout your calculations. Keep up the good work!