The equilibrium constant Kc at 298 K for the reaction A + B C + D is 100. Starting with an equim

Q: Why do we subtract $x$ from reactants and add $x$ to products?

Because the reaction must proceed in the forward direction to reach equilibrium. We determine this by calculating the Reaction Quotient ($Q_c$). Here, $Q_c = (1 \times 1)/(1 \times 1) = 1$. Since $Q_c < K_c$ ($1 < 100$), the reaction proceeds forward, consuming reactants ($-x$) and forming products ($+x$).

Q: Could $x$ be negative?

If $Q_c > K_c$, the reaction would proceed backward, and the change for reactants would be $+x$ and for products $-x$. Here, $Q_c < K_c$, so the forward direction is correct, and $x$ represents a positive amount of substance reacting.

Q: Why did we ignore the negative root when taking the square root?

If we took $\frac{1+x}{1-x} = -10$, we would get $1+x = -10 + 10x \implies 9x = 11 \implies x = 11/9 > 1$. But $x$ cannot be greater than 1 because the initial concentration of reactants is 1 M (we can't consume more than we have). Thus, the negative root is physically impossible.

Q: How do we quickly convert $20/11$ to a decimal?

You can use simple long division. $20 \div 11 = 1$ with a remainder of $9$. Bring down a zero to make it $90$. $90 \div 11 = 8$ with a remainder of $2$. Bring down a zero to make it $20$. $20 \div 11 = 1$, and so on. It repeats as $1.8181...$

Q: Why is the answer multiplied by $10^{-2}$?

This is a common format in numerical answer type questions in JEE to ensure the final answer can be entered as an integer into the computer system. Always read the required format carefully to avoid losing marks on a correct calculation!

Question

The equilibrium constant $K_C$ at 298 K for the reaction $A + B ightleftharpoons C + D$ is 100. Starting with an equimolar solution with concentrations of $A, B, C$ and $D$ all equal to 1M, the equilibrium concentration of $D$ is ......... $	imes 10^{-2}$ M. (Nearest integer)

Sigma Percentile · Accepted Answer

Imagine you are a chemist in a laboratory, carefully mixing chemicals $A$, $B$, $C$, and $D$ into a single reaction vessel. You ensure that the initial concentration of every single species is exactly $1 	ext{ M}$. The reaction is governed by the equation:

$$A + B ightleftharpoons C + D$$

But the system is not at rest. The equilibrium constant $K_C$ for this reaction at $298 	ext{ K}$ is given as $100$. This is a large number, indicating that at equilibrium, the products ($C$ and $D$) will be heavily favored over the reactants ($A$ and $B$).

### The Reaction Quotient ($Q_c$) vs Equilibrium Constant ($K_c$)

Before we dive into calculations, let's ask a fundamental question: Which way will the reaction shift? To answer this, we calculate the Reaction Quotient, $Q_c$, using our initial concentrations:

$$Q_c = \frac{[C]_0[D]_0}{[A]_0[B]_0} = \frac{1 	imes 1}{1 	imes 1} = 1$$

Since $Q_c = 1$ is much less than $K_c = 100$, the reaction must proceed in the **forward direction** to reach equilibrium. This means reactants $A$ and $B$ will be consumed, and products $C$ and $D$ will be formed.

### Setting up the ICE Table

The ICE (Initial, Change, Equilibrium) table is the most powerful tool in a chemist's arsenal for solving equilibrium problems. Let's define $x$ as the change in concentration (in molarity) of the reactants as they move towards equilibrium.

*   **Initial:** $[A] = 1$, $[B] = 1$, $[C] = 1$, $[D] = 1$
*   **Change:** Since the reaction moves forward, $A$ and $B$ decrease by $x$ ($-x$), while $C$ and $D$ increase by $x$ ($+x$).
*   **Equilibrium:** $[A] = 1-x$, $[B] = 1-x$, $[C] = 1+x$, $[D] = 1+x$

### The Master Equation

Now, we substitute these equilibrium concentrations into the expression for the equilibrium constant $K_c$:

$$K_c = \frac{[C][D]}{[A][B]}$$

$$100 = \frac{(1+x)(1+x)}{(1-x)(1-x)}$$

Notice the beautiful mathematical symmetry here. Both the numerator and the denominator are perfect squares. We can rewrite the equation as:

$$100 = \left(\frac{1+x}{1-x}ight)^2$$

Taking the square root of both sides simplifies our lives immensely. The square root of $100$ is $10$.

*(Note: We ignore the negative root, $-10$, because it would yield an $x$ value greater than $1$, which is physically impossible since we only started with $1 	ext{ M}$ of reactants!)*

$$\frac{1+x}{1-x} = 10$$

### Final Calculation

Now, it's just a matter of simple algebra. Cross-multiply to solve for $x$:

$$1 + x = 10(1 - x)$$

$$1 + x = 10 - 10x$$

Bringing all the $x$ terms to one side:

$$11x = 9 \implies x = \frac{9}{11}$$

We are asked to find the equilibrium concentration of $D$, which according to our ICE table is $1 + x$. Let's substitute our value of $x$:

$$[D] = 1 + \frac{9}{11} = \frac{11 + 9}{11} = \frac{20}{11} 	ext{ M}$$

Converting this fraction to a decimal gives us $1.81818... 	ext{ M}$. The question specifically asks for the answer in the format of $... 	imes 10^{-2} 	ext{ M}$, rounded to the nearest integer.

$$1.818 	ext{ M} = 181.8 	imes 10^{-2} 	ext{ M}$$

Rounding $181.8$ to the nearest integer gives us **$182$**.

And there we have it! By systematically applying the principles of chemical equilibrium and leveraging mathematical simplifications, we've arrived at the exact concentration of our product.

Enter Numerical Value:

Visualized Solution

\text{Reaction Setup}

\text{Initial Concentrations}

\text{ICE Table Setup}

\text{Equilibrium Concentrations}

\text{Equilibrium Constant Expression}

\text{Substituting Values}

\text{Simplifying the Equation}

\text{Taking Square Root}

\text{Solving for } x

\text{Finding } x

\text{Concentration of D}

\text{Formatting the Answer}

\text{Conclusion}

The Sigma Insight: Law of Mass Action

The Reaction Quotient ( $Q_{c}$ ) vs Equilibrium Constant ( $K_{c}$ )

Setting up the ICE Table

The Master Equation

Final Calculation

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