VERY STUCK: Second-Order Decomposition Process Total Pressure after 1 hour

Question

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

Answer 1

1/(A) = 1/(A)o + kt

(A) = pressure at time t = ?

(A)_o = initial pressure = 364 mm Hg

k = rate constant = ? but see below for calculation of k

t = time = 1 hour x 3600 sec / hour = 3600 s

First, we'll find k (rate constant) for a 2nd order reaction:

t_1/2 = 1/k(A)_o

410 s = 1 / (k)(364 mm Hg)

k = 6.7x10^-6 mm^-1s^-1

Next, we'll solve for (A)

1/(A) = 1/(A)o + kt

1/(A) = 1(364 mm) + 6.7x10^-6 mm^-1s^-1 x 3600 s

1/(A) = 0.00275 + 0.0241

1(A) = 0.0269

(A) = 37.2 mm Hg This is the pressure of CH3CHO (acetaldehyde) at 1 hour.

The TOTAL pressure will be this value PLUS the pressure the pressure from CH4 and that of CO.

This can be seen in an ICE table as follows:

CH₃CHO <==> CH₄ + CO

364 mm ..............0 mm.....0 mm..........Initial

-x.........................+x.........+x................Change

364-x....................x............x................Equilibrium

364 mm - x = 37.2 and so x = 326.8 mm = 327 mm = pressure CH₄ and the pressure of CO @ equilibrium

In summary we have the following partial pressures @ equilibrium:

(CH₃CHO) = 37.2 mm Hg

(CH₄) = 327 mm Hg

(CO) = 327 mm Hg

TOTAL PRESSURE = 32.7 mm + 327 mm + 327 mm = 687 mm Hg

Answer 2

I don't think the first solution is right, since the formula for the integrated rate law was incorrect. Should've been kt=1/[At]-1/[A0].

Here's my way of solving it (it looks long but the bulk of it is explaining how and why I did each step):

First off, since the decomposition is second order, use the formula as stated iin the beginning.

Then we look at the other known variables: the pressure, temperature are given. It's indicating we should use PV=nRT to transform the concentration (and hence the number of moles) we find from the rate law in to the pressure we're looking for.

With a better idea of how to solve, we look at the rate law again. Another known varaible we haven't used yet is the half life. Plug that in, and we can find k. The equation for half-life in rate law for second order is kt=1/[A0]. Now we need to find [A0]. Use PV=nRT and be wary that the the units have to be in atm, L, mol and K: P=364mmHg=0.4789atm, V=500mL=.5 L, T=518 degrees Celsius=791.15 K and R is the constant 0.0821 atm*L/mol*K. Move RT to the left, and you get from PV=nRT to PV/RT=n. Plug all previously calculated varaibles in, and you get n=3.686*10^-3 mol. Divide that by the volume of 0.5L to get the initial concentration: 7.372*10^-3 M. Transform kt=1/[A0] to k=1/(t*[A0]). You plug in the half-life t=410s and [A0]= 7.372*10^-3 M to find rate constant k=0.3308 M^-1*t^-1.

With the knowledge of what k is, we can try to solve for [At] when t=1 hour=3600 seconds. Plug k, t and [A0] into kt=1/[At]-1/[A0] and you get 0.3308*3600=1/[At]-1/7.372*10^-3. Solve it and you get [At]=7.539*10^-4 M.

Here comes the tricky part: to find the total pressure, you have to account for ALL gases. You have this amount of CH3CHO left, but some of it have already converted to CH4 and CHO, each of which have the same number of moles of the CH3CHO that have decomposed, since the coefficients of all three of those molecules are 1 in the chemical equation. Let's find the moles of CH3CHO left first. The concentration of it is 7.539*10^-4 M in a 0.5 L container, so the number of moles is 3.7695*10^-4 mol. Now we move onto to CH4 and CHO. We just need to find the moles of CH3CHO that have composed and multiply by that by two to find the combined moles of those two products. Looking back, we see that we calculated the initial number of moles of CH3CHO as 3.686*10^-3 mol. Subtract that by the remaining 3.7695*10^-4 mol and you get 3.30905*10^-3 mol. Multiply that by 2 and you get 6.6181*10^-3 mol. Add the previously calculated moles of remaining CH3CHO, which is 3.7695*10^-4 mol, and you get 6.995 * 10^-3 mol.

Whew, we finally have the total number of moles of gases! Just one last step to get to total pressure. Use PV=nRT, with the previously calculated V=.5 L, T=791.15 K and R the constant 0.0821 atm*L/mol*K, we have the answer P=nRT/V=0.909 atm.

I hope this was helpful.

Answer 3

The integrated rate equation for 2nd order is kt = 1/P_A0 - 1/P_A where A is the reactant You are asked to find the pressure after 1 hour given a half-life of 410 seconds. If you take the ration of the equation at t = 3600s and t = 1/2 life, you get the following:

3600s/410s = (1/364T - 1/P_A)/(1/364T - 1/182T) Solve this for P_A T is for Torr

The total Pressure is obtained from P_A0 + (P_A0 - P_A) as a mole of gas is gained for each mole reacted (Pressure is proportional to moles)

Please consider a tutor. Take care.