. The thermal decomposition of dinitrogen pentoxide in the gas phase to give nitrogen dioxide and oxygen is a first order reaction. If the half life at 55 °C is 410 s, what is the value of the rate constant k? If the initial concentration of dinitrogen pentoxide in an expt at 55 °C was 0.200 M what will be the concentration of dinitrogen pentoxide, nitrogen dioxide and oxygen after 4 half-lives?
To find the value of the rate constant, we can use the half-life equation for a first-order reaction:
t1/2 = (0.693 / k)
Given that the half-life at 55 °C is 410 s, we can rearrange the equation to solve for the rate constant:
k = 0.693 / t1/2
k = 0.693 / 410
k ≈ 0.00169 s^-1
Now, let's calculate the concentrations of dinitrogen pentoxide, nitrogen dioxide, and oxygen after 4 half-lives.
After each half-life, the concentration of the reactant is halved. So after 4 half-lives, the concentration will be reduced by a factor of 2^4 = 16.
Initial concentration of dinitrogen pentoxide (N2O5) = 0.200 M
Final concentration of N2O5 = 0.200 M / 16 = 0.0125 M
Since the reaction is a first-order reaction, the ratio of N2O5 to NO2 produced is 1:2. Therefore, the final concentration of nitrogen dioxide (NO2) = 2 * 0.0125 M = 0.025 M
Similarly, for each mole of N2O5 decomposed, one mole of oxygen (O2) is produced. So, the final concentration of oxygen (O2) = 1 * 0.0125 M = 0.0125 M
Hence, after 4 half-lives the concentrations will be:
Dinitrogen pentoxide (N2O5) = 0.0125 M
Nitrogen dioxide (NO2) = 0.025 M
Oxygen (O2) = 0.0125 M
To find the rate constant (k), we can use the formula for the half-life of a first-order reaction:
t1/2 = (0.693 / k)
Given that the half-life (t1/2) at 55 °C is 410 s, we can substitute the values into the formula and solve for k:
410 s = (0.693 / k)
We can rearrange the formula to find k:
k = 0.693 / 410 s
Now let's calculate the value of k:
k = 0.00169 s^-1
Next, we need to determine the concentration of dinitrogen pentoxide (N2O5), nitrogen dioxide (NO2), and oxygen (O2) after 4 half-lives.
The number of half-lives (n) is given by:
n = (time passed) / (half-life)
Since we want to know the concentration after 4 half-lives, n = 4.
To calculate the concentration after 4 half-lives, we can use the formula:
Concentration after n half-lives = Initial concentration * (1/2)^n
Given that the initial concentration of dinitrogen pentoxide is 0.200 M, we can substitute the values into the formula:
Concentration of dinitrogen pentoxide after 4 half-lives = 0.200 M * (1/2)^4
Concentration of dinitrogen pentoxide = 0.200 M * 0.0625
Concentration of dinitrogen pentoxide = 0.0125 M
Similarly, we can calculate the concentration of nitrogen dioxide (NO2) and oxygen (O2) after 4 half-lives using the same formula. Since dinitrogen pentoxide decomposes into an equimolar amount of nitrogen dioxide and oxygen, the concentrations will be the same.
Concentration of nitrogen dioxide after 4 half-lives = Concentration of oxygen after 4 half-lives
= 0.200 M * (1/2)^4
= 0.200 M * 0.0625
= 0.0125 M
Therefore, after 4 half-lives, the concentration of dinitrogen pentoxide, nitrogen dioxide, and oxygen will all be 0.0125 M.