Experiments show that the concentration of substance X in a ﬁrst order chemical reaction at future time (in seconds) t i.e. [X] is described by the following diﬀerential equation:

Question

Experiments show that the concentration of substance X in a ﬁrst order chemical reaction at future time (in seconds) t i.e. [X] is described by the following diﬀerential equation:

d[X]/ dt= .001[X],
ﬁnd:
(a) Find an expression for the concentration [X] after t seconds if the initial concentration is W by solving the diﬀerential equation.

Answer 1

dx/dt = .001x

dx/x = .001 dt
lnx = .001t+c
x(t) = ce^.001t
x(0) = W, so
x(t) = We^.001t

Answer 2

d[X]/dt = (0.001)*[X]

=> d[X]/[X] = (0.001)*dt

Integrating both sides,

∫(1/[X])*d[X] = (0.001)*∫dt
=> log[X] = 0.001*t + C

Now, the question says that when t = 0, X = W

=> log[W] = 0 + C = C

Putting this value of C in the original equation,

log[X] = 0.001*t + log[W]
=> log([X]) - log([W]) = 0.001*t
=> log([X]/[W]) = 0.001*t
=> [X]/[W] = e^(0.001*t)

=> [X] = [W]*e^(0.001*t)

Answer 3

To find an expression for the concentration [X] after t seconds, we need to solve the given differential equation:

d[X]/dt = 0.001[X]

This is a first-order differential equation of the form dy/dt = k*y, where y is the variable of interest ([X] in this case) and k is a constant (0.001 in this case).

To solve this type of differential equation, we can separate the variables and integrate. Let's go through the steps:

1. Start by separating the variables, which means moving all terms involving [X] to one side and terms involving t to the other side:

d[X]/[X] = 0.001*dt

2. Now, integrate both sides of the equation. Integrating d[X]/[X] gives us ln([X]), and integrating 0.001*dt with respect to t gives us 0.001t. The equation becomes:

ln([X]) = 0.001t + C

Here, C is the constant of integration.

3. To find the value of C, we can use the initial concentration [X] = W at t = 0. Substituting these values into the equation:

ln(W) = 0.001*0 + C

ln(W) = C

4. Now, substitute this value of C back into the equation:

ln([X]) = 0.001t + ln(W)

5. To get rid of the natural logarithm, we can exponentiate both sides of the equation:

e^(ln([X])) = e^(0.001t + ln(W))

[X] = e^(0.001t) * e^(ln(W))

6. Simplifying the equation, we know that e^(ln(W)) is just W:

[X] = W * e^(0.001t)

So, the expression for the concentration [X] after t seconds, given an initial concentration W, is [X] = W * e^(0.001t).