[spacebee]

Hi,

I am confused by the 2nd Order reaction rate equations. I recently had a problem where the teacher used a different equation than me to solve a problem where Delta T = ?, Ca0 (initial concentration) = 0.25, Ca (Final Concentration) = 0.125. and K = 0.05 l/mol*min

Teacher used: (1/Ca) - (1/Ca0) = kT
I used: -dCa/dt = kCa^2 or essentially -(Ca - Ca0)/(t2-t1) = kCa^2
These are the equations listed in the NCEES PE reference handbook for the Environmental Engineering exam and they appear to be what other sources have posted as well.

Using the first equation I get t = 80 min, using the second equation I get t = 160 minutes so I derived the equation I used from the teachers equation and I do not get the same equation as above. Observe:

(1/Ca)(Ca0/Ca0) - (1/Ca0)(Ca/Ca) = K(t2-t1)
(Ca0-Ca)/(Cao*Ca) = K(t2-t1)
(Ca0-Ca)/(t2-t1) = KCao*Ca
-(Ca-Ca0)/(t2-t1) = KCa0*Ca

In other words, the right side of the equation is not equal to kCa^2, it is equal to kCa0*Ca. When I use kCa0*Ca (0.125*.25*.05)I get the correct answer of 80 min, when I use (.05*.125^2) I get 160 mins.

What am I not understanding here?

Thanks!

[Corribus]

Can you let us know what the actual question is?

[spacebee]

Yes, the question stated "The following second order reaction occurs in a constant volume batch reactor with a rate of 0.05 L/mol*min. 2A(g) -> 2B(g) How long will it take for half of the original solution to decompose?"

[Corribus]

You sure that's all the information? Half-life of 2nd order reaction depends on the initial concentration, which you have not provided.
(Well, you provided some stuff in your first post, but its relation to your second post isn't clear. Is the initial concentration 0.25 mol/L? You also didn't provide units. Getting problems right begins with being careful with units and laying out all the information at hand in a clear, consistent way.)

[Corribus]

I used: -dCa/dt = kCa^2 or essentially -(Ca - Ca0)/(t2-t1) = kCa^2

Also, just as a general note, this is not how you solve differential equations. It is no surprise you get the wrong answer doing it this way. I.e., there is no "essentially" - the first equation does not translate to the second one. If you would like me to show you how you derive the correct one (the one your teacher used), I am happy to do that.

[spacebee]

Good point, I had not noticed that since the solution provided had the concentrations listed. I have uploaded the problem statement and solution.

[spacebee]

I apologize, I have been out of school for some time and unfortunately as as hydraulic engineer a lot of what I learned in school is not something I really use anymore so it has more or less left my brain. But, there is no exam for hydraulic engineers and thus I must revisit these old topics to get my license. Please, do show me the correct way to integrate the equation. My guess is that you will get the equation the professor used and I should just steer clear of the equations with derivatives in them when I take the exam less I find myself integrating equations where the integrated equation is already provided.

[Corribus]

No problem. Your basic error (don't feel too bad, a lot of people make it) is that if y is some variable, dy≠Δy. In Calculus, dy means the change in the y variable in the limit that the interval becomes infinitely small. Δy usually refers to the change over a large interval (final value minus initial value). In most cases, you can't just swap in Δy for dy and Δx for dx because if you do, you're not treating the derivative of the function as... well, a derivative.

Here is how you derive the proper equation to solve the problem. I will assume you understand where this differential equation comes from:

[tex]-\frac{dA}{dt}=kA^2[/tex]

First step is to convert the equation to:

[tex]\frac{dA}{A^2}=-k dt[/tex]

Integrate both side:

[tex]\int \frac{dA}{A^2}=\int -k dt[/tex]

Perform integration:

[tex]-\frac{1}{A} + C =-kt[/tex]

Here, C is a constant of integration. We can put it on either side, doesn't matter. To determine C, we use the initial condition that at t = 0, the concentration is A₀. Doing a little algebra, you can show that

[tex]C=\frac{1}{A_0}[/tex]

So (after more algebra)

[tex]\frac{1}{A} = \frac{1}{A_0} + kt[/tex]

That's the equation your teacher used. You can take this a pinch further and derive an equation for the half life by setting A to A₀/2 and solving for t, which gives

[tex]t_{\frac{1}{2}}= \frac{1}{kA_0}[/tex]

So, if you put in 0.25 M into this equation, and k = 0.05 L/mol min, you get 80 min.

Hope that helps.

[spacebee]

Yes this makes sense. The problem was fake calculus Thank you for explaining.

A little algebra with the equation you derived matches the equation I used when I was getting the correct answer...