Animated Solution for Chemistry - Chemical Equilibrium: The equilibrium constants $K_{p_1}$ and $K_{p_2}$ for the reactions $X\rightleftharpoons 2Y$ and $Z\rightleftharpoons P+Q$, respectively are in the ratio of $1:9$. If the degree of dissociation of $X$ and $Z$ be equal, then the ratio of total pressure at these equilibria is

Imagine two separate reaction vessels, each hosting a delicate chemical dance. In the first vessel, a gas $X$ is breaking apart into two molecules of $Y$. In the second vessel, a gas $Z$ is splitting into two different molecules, $P$ and $Q$.

We are given a fascinating piece of information: the equilibrium constant of the first reaction ($K_{p_1}$) is exactly one-ninth of the equilibrium constant of the second reaction ($K_{p_2}$). Furthermore, the degree of dissociation, $\alpha$, is identical for both gases. Our mission is to find the ratio of the total pressures, $p_1$ and $p_2$, in these two vessels.

Let's dive into the mathematics of the first reaction: $X\rightleftharpoons 2Y$.

To keep things simple, we assume we start with exactly $1$ mole of $X$. At equilibrium, a fraction $\alpha$ of it has dissociated. This leaves us with $(1-\alpha )$ moles of $X$. Since each molecule of $X$ produces two molecules of $Y$, we generate $2\alpha$ moles of $Y$.

The total number of moles in the vessel is the sum of these: $(1-\alpha )+2\alpha =1+\alpha$.

Using Dalton's Law, the partial pressure of a gas is its mole fraction multiplied by the total pressure $p_1$. $p_X=\left(\frac{1-\alpha }{1+\alpha }\right)p_1$ $p_Y=\left(\frac{2\alpha }{1+\alpha }\right)p_1$

Now, we construct the equilibrium constant $K_{p_1}$:

Simplifying this algebraic fraction, one power of $p_1$ cancels out, and the denominator elegantly collapses into a difference of squares:

Now, let's shift our focus to the second reaction: $Z\rightleftharpoons P+Q$.

Again, starting with $1$ mole of $Z$, at equilibrium we have $(1-\alpha )$ moles of $Z$. This time, it produces $1$ mole of $P$ and $1$ mole of $Q$ for every mole dissociated, giving us $\alpha$ moles of $P$ and $\alpha$ moles of $Q$.

Interestingly, the total number of moles is exactly the same as in the first vessel: $(1-\alpha )+\alpha +\alpha =1+\alpha$.

The partial pressures, based on the total pressure $p_2$, are: $p_Z=\left(\frac{1-\alpha }{1+\alpha }\right)p_2$ $p_P=p_Q=\left(\frac{\alpha }{1+\alpha }\right)p_2$

Constructing $K_{p_2}$:

Simplifying this yields a remarkably similar expression:

Many textbooks will tell you to assume that $\alpha$ is very small, meaning ${\alpha }^2\ll 1$, allowing you to ignore the denominator. But look closely—we don't need to approximate anything!

We are given the ratio:

Let's substitute our exact, unapproximated expressions:

Notice the sheer mathematical beauty here. The complex $(1-{\alpha }^2)$ denominators perfectly cancel each other out. The ${\alpha }^2$ terms in the numerators also vanish. We are left with a pristine, simple equation:

Solving for the ratio of the pressures, we get:

$$\frac{p_1}{p_2}=\frac{1}{36}$$

This problem is a masterclass in trusting the algebra. By carrying the exact expressions through to the final ratio, we avoided unnecessary approximations and arrived at the exact answer with elegance.