In a chemical reaction, A + 2B 2C + D, the initial concentration of B was 1.5 times of the conce

Q: Why do we subtract $2x$ from the concentration of B instead of just $x$?

The change in concentration is directly proportional to the stoichiometric coefficients in the balanced chemical equation. Since the coefficient of B is 2, it means 2 moles of B are consumed for every 1 mole of A. Therefore, if A decreases by $x$, B must decrease by $2x$.

Q: Can I assume the initial concentration of A is 1 M instead of $a_0$?

Yes, absolutely! Since the final equilibrium constant $K$ is a ratio where the initial concentration terms cancel out completely, assuming $a_0 = 1$ M (or any other arbitrary number) will yield the exact same result and might make the algebra feel less abstract.

Q: Why is the equilibrium constant $K$ unitless in this specific problem?

The units of $K$ depend on the change in the number of moles of gas ($\Delta n_g$). Here, $\Delta n_g = (\text{moles of products}) - (\text{moles of reactants}) = (2 + 1) - (1 + 2) = 0$. When $\Delta n_g = 0$, all concentration units cancel out, leaving $K$ as a dimensionless number.

Q: What if the problem stated that the equilibrium concentration of C was equal to A?

You would follow the exact same ICE table setup. However, your constraint equation would change. Instead of equating $[A]$ and $[B]$, you would set $[A]_{eq} = [C]_{eq}$, which means $a_0 - x = 2x$. You would then solve for $x$ ($x = a_0/3$) and proceed with the calculation.

Q: I got confused when squaring the denominator. How did $(0.5a_0)^2$ become $0.25a_0^2$?

Remember the exponent rule $(ab)^n = a^n b^n$. When you square the term $(0.5a_0)$, you must square both the numerical coefficient and the variable. So, $(0.5)^2 = 0.25$ and $(a_0)^2 = a_0^2$, resulting in $0.25a_0^2$.

Question

In a chemical reaction,
$A + 2B \rightleftharpoons 2C + D$, the initial concentration of $B$ was 1.5 times of the concentration of $A$, but the equilibrium concentrations of $A$ and $B$ were found to be equal. The equilibrium constant ($K$) for the aforesaid chemical reaction is

Options:

(A) $\frac{1}{4}$

(B) 16

(C) 1

(D) 4

Sigma Percentile · Accepted Answer

The journey to mastering chemical equilibrium often begins with a single, powerful tool: the **ICE table**. In this problem, we are tasked with finding the equilibrium constant $K$ for a reaction, given a specific relationship between the initial and equilibrium concentrations of the reactants. Let's dive into the step-by-step logic that unravels this mystery.

### Setting the Stage: The ICE Table

The reaction is given as:
$$A + 2B \rightleftharpoons 2C + D$$

We are told that the initial concentration of $B$ is 1.5 times that of $A$. To make our algebra clean, let's assume the initial concentration of $A$ is $a_0$. Consequently, the initial concentration of $B$ becomes $1.5a_0$. Since the reaction hasn't started, the products $C$ and $D$ are at $0$.

Now, we introduce the "Change" row. Let $x$ be the amount of $A$ that reacts to reach equilibrium. According to the stoichiometry of the balanced equation, for every 1 mole of $A$ that reacts, 2 moles of $B$ must also react. Therefore, the change for $B$ is $-2x$. On the product side, 2 moles of $C$ and 1 mole of $D$ are formed, making their changes $+2x$ and $+x$, respectively.

Adding the initial concentrations and the changes gives us the equilibrium concentrations:
*   $[A]_{eq} = a_0 - x$
*   $[B]_{eq} = 1.5a_0 - 2x$
*   $[C]_{eq} = 2x$
*   $[D]_{eq} = x$

### The Master Constraint

The problem provides a crucial piece of information: at equilibrium, the concentrations of $A$ and $B$ are equal. This is our master constraint, the key to unlocking the value of $x$.

$$[A]_{eq} = [B]_{eq}$$
$$a_0 - x = 1.5a_0 - 2x$$

Solving this linear equation is straightforward. By rearranging the terms, we get:
$$2x - x = 1.5a_0 - a_0$$
$$x = 0.5a_0$$

This tells us that exactly half of the initial amount of $A$ has reacted.

### Calculating the Equilibrium Constant

With $x$ found in terms of $a_0$, we can now determine the exact equilibrium concentrations for all species:
*   $[A] = a_0 - 0.5a_0 = 0.5a_0$
*   $[B] = 1.5a_0 - 2(0.5a_0) = 0.5a_0$
*   $[C] = 2(0.5a_0) = a_0$
*   $[D] = 0.5a_0$

The equilibrium constant $K$ is defined by the Law of Mass Action as the ratio of the product concentrations to the reactant concentrations, each raised to the power of their stoichiometric coefficients:
$$K = \frac{[C]^2[D]}{[A][B]^2}$$

Substituting our equilibrium values into this expression:
$$K = \frac{(a_0)^2 (0.5a_0)}{(0.5a_0) (0.5a_0)^2}$$

Notice how elegantly the terms cancel out. The $(0.5a_0)$ in the numerator and denominator cancel immediately:
$$K = \frac{a_0^2}{(0.5a_0)^2}$$
$$K = \frac{a_0^2}{0.25a_0^2}$$

The $a_0^2$ terms cancel out, leaving us with a purely numerical value:
$$K = \frac{1}{0.25} = 4$$

**The equilibrium constant for this reaction is 4.** This problem beautifully illustrates how setting up a systematic ICE table and carefully applying the given constraints can simplify seemingly complex equilibrium scenarios into basic algebra.

Select Answer:

Visualized Solution

The Reaction Setup

Initial Concentrations

Change in Concentration

Equilibrium Concentrations

Applying the Constraint

Solving for $x$

Actual Equilibrium Values

Equilibrium Constant $K$

Calculating $K$

Conclusion

The Sigma Insight: Law of Mass Action

Setting the Stage

The Master Constraint

Calculating the Equilibrium Constant

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