# Solving an ODE in the context of a chemical reaction

Two chemicals $$\mathrm{X}$$ and $$\mathrm{Y}$$ react in such a way that $$1\,\rm g$$ of $$\mathrm{X}$$ combines with $$4\,\rm g$$ of $$\mathrm{Y}$$ to produce a compound $$\mathrm{Z}$$ which is written as:

$$\mathrm{X + 4Y \to Z}$$

The rate at which $$\mathrm{X}$$ and $$\mathrm{Y}$$ react (the rate at which $$\mathrm{Z}$$ is formed) is proportional to the product of the amount of $$\mathrm{X}$$ and $$\mathrm{Y}$$ that remained untransformed at that time. Assume there is initally $$50\,\rm g$$ of $$\mathrm{X}$$ and $$32\,\rm g$$ of $$\mathrm{Y}$$, and it is known that $$30\,\rm g$$ of $$\mathrm{Z}$$ are formed in $$10\,\rm min$$.

Derive a differential equation initial value problem (sic) for $$Z(t)$$.

Right now, I have this, where $$\alpha (t)$$ represents the transformed $$\mathrm{X}$$ at time $$t$$, and $$\beta (t)$$ represents the transformed $$\mathrm{Y}$$ at time $$t$$:

$$\frac{\Bbb dZ}{\Bbb dt} = k\big(50 - \alpha (t)\big) \big(32 - \beta (t)\big)$$

Where do I go from here? I know I'll have to use the given relation, and eventually have to only use a $$Z$$ instead of $$\alpha$$ and $$\beta$$, but how?

Once I get a proper separable equation, I should be good to go.

• Thanks for the edits, I wasn't really sure how often to use LaTeX. Seems like whenever possible, good to know! – Tetramputechture Sep 9 '15 at 21:18

Let $x$ be the quantity of $\mathrm{Z}$ that was formed. So, the remaining quantities of each component are (I assume you know this as it pure chemistry):

\begin{align} n_\mathrm{X}&=50-x\\ n_\mathrm{Y}&=32-4x\\ n_\mathrm{Z}&=x \end{align}

And, since the rate at which $\mathrm{Z}$ reacts is proportional to the quantities you mentioned, then:

$$\frac{dn_\mathrm{Z}}{dt}=\frac{dx}{dt}=k(50-x)(32-4x)$$

I assume that's easy and straightforward. You can now separate it and solve for $x$.

• Alright, that seems quite straightforward to me, which is why I was unsure when I started tackling the problem (I remember my professor taking more steps). Thank you! – Tetramputechture Sep 9 '15 at 21:24
• @Tetramputechture You're welcome. Glad I helped – Oussama Boussif Sep 9 '15 at 21:27
• Apparently the "correct" DE was dx/dt = k(50 - (x/5)(32 - (4x/5)). I just hope I don't get docked too many points for solving that wrong equation, too. :( – Tetramputechture Sep 15 '15 at 15:43