When acetaldehyde at a pressure of 364 mmHg is introduced into an evacuated 500. mL flask at 518^$\circ$C, the half-life for the second-order decomposition process, CH₃CHO $\rarr$CH₄ + CO, is 410. s. What will the total pressure in the flask be after 1.00 hour

Question

Quizplus · Accepted Answer

First, we need to convert the temperature to Kelvin:

518F1F1F1S1F1F1F10C = (518+273)K = 791K

To find the rate constant for the reaction, we can use the half-life formula for a second-order reaction:

t1/2 = 1 / (k [A]0)

where t1/2 = 410 s, [A]0 is the initial concentration of acetaldehyde (which we can assume to be equal to the equilibrium concentration since the flask was evacuated), and k is the rate constant.

Solving for k:

k = 1 / (t1/2 [A]0) = 1 / (410 s * (364 mmHg / 760 mmHg)) = 3.09 x 10^-4 M^-1 s^-1

Next, we can use the integrated rate law for a second-order reaction:

1 / [A]t = kt + 1 / [A]0

where [A]t is the concentration of acetaldehyde at time t, [A]0 is the initial concentration (which we can assume to be equal to the equilibrium concentration), and k is the rate constant.

We want to find the total pressure in the flask after 1 hour (3600 s). At this point, the concentration of acetaldehyde will have decreased, but we can assume that the volume of the flask remains constant (since it's a closed system) and that the CO product doesn't contribute significantly to the total pressure. Therefore, we can use the ideal gas law:

Ptotal = nRT / V

where n is the number of moles of gas, R is the gas constant, and T and V are the temperature and volume of the flask, respectively.

We can determine the number of moles of acetaldehyde from its concentration:

[A]t = n / V

n = [A]t * V

where V is in liters, [A]t is in moles/L, and n is in moles.

We can use the integrated rate law to find [A]t:

1 / [A]t = kt + 1 / [A]0

1 / [A]t = (3.09 x 10^-4 M^-1 s^-1) (3600 s) + 1 / (364/760 M)

[A]t = 0.000175 M

Substituting into the expression for n:

n = (0.000175 M) (0.5 L) = 8.75 x 10^-5 mol

Finally, we can substitute all of our values into the ideal gas law:

Ptotal = (8.75 x 10^-5 mol) (0.08206 L atm/mol K) (791 K) / (0.5 L)

Ptotal = 691 mmHg

Therefore, the answer is choice D.

When Acetaldehyde at a Pressure of 364 MmHg Is Introduced ∘\circ∘

When Acetaldehyde at a Pressure of 364 MmHg Is Introduced $\circ$