. 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?

Question

. 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?

Answer 1

Let the initial and final concentrations be Co and Ct.

To find k:
ln(Ct/Co)=-kt
ln(0.5) = -k(410s)
k = ____?

To find the concentration at (4)(410s), use the same equation:
ln(Ct/Co)=-kt
Use the value of k from the first part, and t = 4x410=1640s, then solve for for Ct.

Answer 2

To find the value of the rate constant (k) for the thermal decomposition of dinitrogen pentoxide, we can use the half-life (t1/2) and the formula for first-order reactions:

t1/2 = ln(2) / k

Rearranging the formula, we can solve for k:

k = ln(2) / t1/2

Plugging in the given half-life, we can calculate the rate constant:

k = ln(2) / 410 s

Now, let's calculate the value of k using this formula:

k = 0.693 / 410 s

k ≈ 0.00169 s^(-1)

Now, let's move on to the second part of the question. If the initial concentration of dinitrogen pentoxide is 0.200 M and the reaction has a first-order rate constant, we can use the equation for a first-order reaction to determine the concentrations after a certain number of half-lives:

[Reactant] = [Reactant]₀ * 2^(-n)

where [Reactant] is the final concentration, [Reactant]₀ is the initial concentration, and 'n' is the number of half-lives.

Given that we want to find the concentrations after 4 half-lives, we will substitute n = 4 in the equation.

For dinitrogen pentoxide:

[DNPO] = [DNPO]₀ * 2^(-4)

[DNPO] = 0.200 M * 2^(-4)

[DNPO] = 0.200 M * 0.0625

[DNPO] = 0.0125 M

For nitrogen dioxide (NO₂):

[NO₂] = [NO₂]₀ * (1 - 2^(-n))

[NO₂] = 0.200 M * (1 - 2^(-4))

[NO₂] = 0.200 M * (1 - 0.0625)

[NO₂] = 0.200 M * 0.9375

[NO₂] = 0.1875 M

For oxygen (O₂):

[O₂] = [O₂]₀ - [DNPO]

[O₂] = 0.200 M - 0.0125 M

[O₂] = 0.1875 M

After 4 half-lives, the concentration of dinitrogen pentoxide (DNPO) is 0.0125 M, the concentration of nitrogen dioxide (NO₂) is 0.1875 M, and the concentration of oxygen (O₂) is also 0.1875 M.