The rate of the following reaction:
2 NO2 2 NO + O2,
is characterized by the rate constant k = 0.775 L·mol-1s-1. What is the half-life for this reaction at the same conditions when the initial concentrations are 0.0115 M? (Round your answer to 3 significant figures.)
Hint: The units of the rate constant are linked with the order of reaction.
Here is a site that lists the units of the order. Scroll about half way down the page to see this is a second order equation. Then look in your text for the integrated form (the site also shows the formula), substitute the numbers from the problem and solve. The site even has the NO2 problem (but not your numbers) as an example.
http://en.wikipedia.org/wiki/Rate_equation
To determine the half-life of a reaction, we first need to determine the order of the reaction. The order of reaction can be determined from the units of the rate constant (k).
In this case, the rate constant (k) is given as 0.775 L·mol^(-1)·s^(-1).
For a reaction with the general form: A → products, the rate expression can be written as: Rate = k [A]^n, where [A] represents the concentration of reactant A and n represents the order of reaction.
Comparing the units of the rate constant (k) with the units of concentration ([A])^n, we can determine the order of reaction (n).
In this case, the units of the rate constant (k) are in L·mol^(-1)·s^(-1), which is consistent with a second-order reaction (since the sum of the exponents is 1 + 1 = 2).
Therefore, the given reaction is a second-order reaction.
Next, we can use the equation for the half-life of a second-order reaction to calculate the half-life (t1/2). The equation is:
t1/2 = 1 / (k [A]0),
where [A]0 represents the initial concentration of the reactant.
Plugging in the given values, we have:
t1/2 = 1 / (0.775 L·mol^(-1)·s^(-1) * 0.0115 M).
Calculating this expression gives:
t1/2 = 1 / (0.00892625 L·mol^(-1)·s^(-1)).
t1/2 ≈ 112.080 s.
Rounded to three significant figures, the half-life of the reaction is approximately 112 s.