For a reaction, X + Y ⇌ 2Z
1.0 mol of X, 1.5 mol of Y, and 0.5 mol of Z, were taken in a 1 L vessel

Q: Why did we take the change in Z as +2a?

The balanced chemical equation is $X + Y \rightleftharpoons 2Z$. For every $1\text{ mole}$ of $X$ that reacts, $2\text{ moles}$ of $Z$ are produced. Hence, the change is $+2a$.

Q: Does the initial presence of Z affect the direction of the reaction?

It could! To be absolutely sure, we would calculate the reaction quotient $Q_c$. However, since the equilibrium concentration of $Z$ ($1.0\text{ M}$) is greater than its initial concentration ($0.5\text{ M}$), we know the reaction must have proceeded in the forward direction.

Q: Why did we convert decimals to fractions in the final calculation?

Converting decimals like $0.75$ to $\frac{3}{4}$ and $1.25$ to $\frac{5}{4}$ makes manual multiplication and division much faster and less prone to silly mistakes, especially in exams where calculators are not allowed.

Q: What if the volume of the vessel was not 1 L?

If the volume was different, say $V$ liters, we would have to divide the initial moles by $V$ to get the initial molarity before setting up the ICE table. Here, dividing by $1$ kept the numerical values the same.

Q: Why does $K_{eq}$ not have any units in this problem?

The unit of $K_{eq}$ is $(\text{mol L}^{-1})^{\Delta n_g}$. Here, $\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants}) = 2 - (1 + 1) = 0$. Therefore, $K_{eq}$ is unitless.

Question

For a reaction, $X + Y ightleftharpoons 2Z$
$1.0	ext{ mol}$ of $X$, $1.5	ext{ mol}$ of $Y$, and $0.5	ext{ mol}$ of $Z$, were taken in a $1	ext{ L}$ vessel and allowed to react. At equilibrium, the concentration of $Z$ was $1.0	ext{ mol L}^{-1}$. The equilibrium constant of the reaction is $\frac{x}{15}$. The value of $x$ is ......... .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are a chemist setting up a reaction in a perfectly sized $1	ext{ L}$ flask. You carefully measure out $1.0	ext{ mol}$ of gas $X$, $1.5	ext{ mol}$ of gas $Y$, and $0.5	ext{ mol}$ of gas $Z$.

Because the volume of our vessel is exactly $1	ext{ L}$, the number of moles directly gives us the molar concentration. This makes our calculations incredibly straightforward!

We are dealing with the reversible reaction:
$$X + Y ightleftharpoons 2Z$$

To keep track of how the concentrations change as the system reaches equilibrium, we will use an ICE (Initial, Change, Equilibrium) table. This is one of the most powerful tools in a chemist's arsenal.

### The Master Equation

Let's assume that as the reaction proceeds forward, $a	ext{ moles}$ of $X$ and $Y$ are consumed.

According to the stoichiometry of the balanced equation, for every $1	ext{ mole}$ of $X$ that reacts, $2	ext{ moles}$ of $Z$ are produced. Therefore, the change in concentration for $Z$ will be $+2a$.

At equilibrium, the concentrations will be the sum of their initial values and the changes:
$$[X] = 1.0 - a$$
$$[Y] = 1.5 - a$$
$$[Z] = 0.5 + 2a$$

### Unlocking the Extent of Reaction

Here is where the problem gives us a crucial piece of information. We are told that the equilibrium concentration of $Z$ is exactly $1.0	ext{ M}$.

We can set up a simple algebraic equation:
$$0.5 + 2a = 1.0$$

Solving for $a$, we subtract $0.5$ from both sides to get $2a = 0.5$, which means $a = 0.25	ext{ M}$.

Now that we have unlocked the value of $a$, we can easily find the equilibrium concentrations of our reactants by substituting it back:
$$[X] = 1.0 - 0.25 = 0.75	ext{ M}$$
$$[Y] = 1.5 - 0.25 = 1.25	ext{ M}$$

### Final Calculation

With all the equilibrium concentrations in hand, we can write the expression for the equilibrium constant, $K_{eq}$.

Remember, it is the product concentrations raised to their stoichiometric coefficients divided by the reactant concentrations:
$$K_{eq} = \frac{[Z]^2}{[X][Y]}$$

Let's substitute our values into this expression:
$$K_{eq} = \frac{(1.0)^2}{(0.75)(1.25)}$$

To avoid messy decimal multiplication, **convert the decimals to fractions**. This is a pro-tip for JEE and NEET exams! $0.75$ is $\frac{3}{4}$ and $1.25$ is $\frac{5}{4}$.
$$K_{eq} = \frac{1}{\frac{3}{4} 	imes \frac{5}{4}} = \frac{1}{\frac{15}{16}} = \frac{16}{15}$$

The problem states that $K_{eq} = \frac{x}{15}$. By comparing this with our calculated value, it is crystal clear that:
$$x = 16$$

And there we have it! A beautiful, logical progression from an initial state to a final, satisfying integer answer.

Enter Numerical Value:

Visualized Solution

$X + Y ⇌ 2 Z$

Change in Concentration

Equilibrium Concentrations

Finding $a$

Equilibrium Concentrations of $X$ and $Y$

Equilibrium Constant $K_{e q}$

Substituting Values

Calculating $K_{e q}$

Comparing with Given $K_{e q}$

What if Volume was $2 L$ ?

The Sigma Insight: Law of Mass Action

Analyzing the Setup

The Master Equation

Unlocking the Extent of Reaction

Final Calculation

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