The problem, which I found in a popular book for Physical Chemistry, written by Neeraj Kumar, is fairly simple, being simply 2 equations and 2 variables, yet its simplification is a nightmare.

$\pu{5 atm}$ $\ce{AB(g)}$ and $\pu{0.5 atm}$ $\ce{BC(g)}$ are introduced in a reaction vessel and the following equilibriums are established: 

$$\ce{AB(g) <=> A(g) + B(g)} \qquad K = \pu{10^{-7} atm}$$

$$\ce{BC(g) <=> B(g) + C(g)} \qquad K = \pu{10^{-6} atm}$$

If the equilibrium partial pressure exerted by $\ce{B(g)}$ is around $10^{–n}$, determine the value of $n$.

We start of by writing the simultaneous equilibriums

enter image description here

$$10^{-7} = \dfrac{x(x+y)}{5-x}\tag{...1}$$

$$10^{-6} = \dfrac{y(x+y)}{0.5-y}\tag{...2}$$

Dividing the 2 gives us:

$$10^{-1} = \dfrac{x(0.5-y)}{y(5-x)}$$

This further simplifies to:


We now have $y$ in terms of $x$, so we can substitute $\mathrm{(3)}$ in either $\mathrm{(1)}$ or $\mathrm{(2)}$, but substituting in $\mathrm{(1)}$ gives an easier simplification.

Substituting $\mathrm{(3)}$ in $\mathrm{(1)}$:

$$10^{-7} = \dfrac{x(x+\cfrac{5x}{5+9x})}{5-x}$$

This simplifies to give us:

$$10^{-7} = \dfrac{x(9x^{2}+10x)}{(5+9x)(5-x)}$$

It was the simplification of this cubic equation which gave me a problem.

Generally, approximations can be taken in such equations (by simply looking at the value of $K$, we know that $x$ will be a very small number) but here I don't see any. We can neglect $x$ in front of 5 in the $(5-x)$ term, but I doubt that makes the simplification of the cubic equation any easier.

On plugging the equation into Wolfram Alpha, it gives the following results which can be viewed here.

These are the results:

enter image description here

$x \approx 5 \times 10^{-4}$ is the only valid result in this case.

On plugging this value of $x$ into (3), we get:

y = $\dfrac{25 \times 10^{-4}}{5+9 \times 5 \times 10^{-4}}$ (we neglect $45 \times 10^{-4}$ in front of 5)

$$\implies y = 5 \times 10^{-4}$$

Therefore, the partial pressure of $\ce{B(g)}$ at equilibrium is $= x+y = 2 \times 5 \times 10^{-4} = 10^{-3}$

My only qualms are in solving the cubic we had initially:

$$10^{-7} = \dfrac{x(9x^{2}+10x)}{(5+9x)(5-x)}$$

Without Wolfram Alpha, how would we solve this cubic equation? Approximating $(5-x) \approx 5$ doesn't seem to help our case much. Is there any way to solve this without using a computer, simply using pen and paper?

P.S.: This problem was actually asked in 2013 to high school students in a written competitive exam called JEE, and in that exam, the students only have 5 to 6 minutes for each problem, so imagine the sheer skill required to solve such a difficult cubic equation and get the answer in that time-frame.

  • $\begingroup$ math.utah.edu/~wortman/1060text-tcf.pdf $\endgroup$ Dec 25, 2023 at 15:20
  • $\begingroup$ Usually when the calculations are too tough JEE awards it as a bonus. One more thing, you have given only JEE and not specified whether it is Main or Advanced. $\endgroup$ Dec 25, 2023 at 15:26
  • $\begingroup$ @HarikrishnanM I believe it was the the first year JEE Advanced was conducted. $\endgroup$
    – Bongo Man
    Dec 25, 2023 at 15:28
  • $\begingroup$ Now coming to your question: I tried searching the archives on the JEE Adv 2024 website but could find none. jeeadv.ac.in/archive.html . It would be great if you could post the problem image. $\endgroup$ Dec 25, 2023 at 15:34

1 Answer 1


A relatively straightforward solution is to start with your two equations:



In both cases, the variable in the denominator is negligible relative to the constant, so we simplify both:



If we divide both sides of the second equation by 10 to make the denominator an integer, we see that both equations are identical except for replacement of $x$ with $y$, which tells us that $x=y$. That allows us to substitute $x$ for $y$ to yield

$$10^{−7}=\frac{x(x+x)}{5}$$ $$\implies 5\times 10^{-7} = 2x^2$$ $$\implies x^2=25\times 10^{-8}$$ $$\implies x = 5\times 10^{-4}$$ $$\implies x+y=10^{-3}$$ $$\implies n=3$$

  • 2
    $\begingroup$ Nice! I've never known a book problem to require solving a cubic equation. There always seems to be some simplification that reduces the problem to a quadratic at most. $\endgroup$
    – MaxW
    Dec 26, 2023 at 4:51

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