Differential Equation Rate of decay problem

[clope023]

Homework Statement

A chemical reaction converts a certain chemical into another chemical, and the rate at which the 1st chemical is converted is proportional to the amount of of this chemical present at any time. At the end of 1 hour, 50g of the 1st chemical remain; while at the end of 3 hours only 25g remain.

a) how many grams of the 1st chemical were present initially?

Homework Equations

dx/dt = -kx

x(0) = x0
x(1) = 50g
x(3) = 25g

int[dx/x = -kdt] -> ln(x) = -kt + c

x = ce^(-kt)

x(0) = x0 -> x0 = C(1), x = x0e^(-kt)

The Attempt at a Solution

I'm pretty much stuck where I ended the relevant equations, I'm not sure what to do to get x0 by itself, I tried plugging in the 1st hour value which got me

50 = x0e^(-k), but I need to solve for k or the value of e^(-k) before I can solve for x0, which I could do easily by taking e^(-k)/50 and then I could solve the rest of the problem; any help would be greatly appreciated.

oh and if th3plan is reading this, thanks for the help in the other thread I made.

 

[mighty2000]

You are on the right track with your equations and attempts at solving for the initial amount of the first chemical. To solve for the initial amount, we need to use the given information about the amount of chemical remaining at the end of 1 hour and 3 hours.

From the given information, we know that x(1) = 50g and x(3) = 25g. Plugging these values into the equation x = x0e^(-kt), we get:

50 = x0e^(-k) and 25 = x0e^(-3k)

Now, we can divide the second equation by the first equation to eliminate x0:

25/50 = x0e^(-3k)/x0e^(-k)

1/2 = e^(-2k)

Taking the natural log of both sides, we get:

ln(1/2) = -2k

Solving for k, we get:

k = -ln(1/2)/2 ≈ 0.3466

Now, we can plug this value of k back into the first equation to solve for x0:

50 = x0e^(-0.3466)

x0 = 50/e^(-0.3466) ≈ 60.24g

Therefore, the initial amount of the first chemical was approximately 60.24 grams.

I hope this helps! Keep up the good work in your studies.