# Writing the differential equation

1. Oct 9, 2008

### A_lilah

1. The problem statement, all variables and given/known data

A chemical reaction A -> B is carried out in a closed vessel. If the tank volume is 30.5 gallons, and there is now B in the tank at t=0, how much B in grams does the tank contain after 2 hours? The reaction rate is constant, k, and k= 9.3x10^-3 g/L per minute

2. Relevant equations

B = amount of B in grams, in the vessel at a given time, t (in hours)
dB/dt = inflow of B - outflow of B

3. The attempt at a solution

1. There is now outflow of B because the vessel is closed and the reaction only goes one way, so dB/dt = inflow of B

2. Covert gallons to liters: 30.5gal (1000L / 264.17 gal) = 115.4559564L

3. B (grams)/ 115.4559564 L = amount of B in the vessel at any specific time

4. (B/155.4559564)(-9.3x10^-3) = B * -8.05501967x10^-5 ( units = g^2min / L^2)

****This is where I got a little confused, because the units are so weird, but I integrated anyways, just to see if my solution would make sense******

5. dB/dt = B * -8.05501967x10^-5

6. 1/B dB = -8.05501967x10^-5 dt

7. lnB = -8.05501967x10^-5 *t + C

8. e^(lnB) = e^(-8.05501967x10^-5 *t + C)

9. B = e^(-8.05501967x10^-5 *t + C), or, because this = e^(-8.05501967x10^-5 *t) * e^C, which is just another constant, C, you get:

B = Ce^(-8.05501967x10^-5 *t)
At t= 0, B=0 (from above) --- plugging that in, you find C = 0....

And this is where I am completely lost. We just learned how to do this yesterday,and I got this problem in another class, so any help would be great!
Thanks

Last edited: Oct 9, 2008
2. Oct 9, 2008

### A_lilah

Did I add the reaction rate in right?