The equilibrium constant at 298 K for a reaction, A + B C + D is 100. If the initial concentrati

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. Even though products are present initially, the high value of $K_{eq} = 100$ suggests the equilibrium heavily favors the products, so more reactants will convert into products.

Q: What if I assumed the reaction goes backward?

If you assumed it goes backward, you would add $x$ to reactants and subtract $x$ from products. You would end up with a negative value for $x$, but the final equilibrium concentrations would still be exactly the same!

Q: Why didn't we use the quadratic formula?

We could have! Expanding the squares would give a quadratic equation. However, recognizing that both the numerator and denominator are perfect squares allows us to take the square root of both sides, saving a lot of time and reducing the chance of calculation errors.

Q: Does the volume of the container matter here?

In this specific reaction, the number of moles of gaseous reactants equals the number of moles of gaseous products ($\Delta n_g = 0$). Therefore, the volume terms cancel out in the equilibrium constant expression, making it independent of volume.

Q: Why is the initial concentration of products not zero?

The problem explicitly states that the initial concentrations of *all four species* were 1 M. This is a specific initial condition set by the examiner to test your ability to handle non-standard starting states.

Question

The equilibrium constant at $298	ext{ K}$ for a reaction, $A + B ightleftharpoons C + D$ is $100$. If the initial concentrations of all the four species were $1	ext{ M}$ each, then equilibrium concentration of $D$ (in $	ext{mol L}^{-1}$) will be

Options:

(A) 0.818

(B) 1.818

(C) 1.182

(D) 0.182

Sigma Percentile · Accepted Answer

### Setting the Stage: The ICE Table

Imagine you are setting up a chemical reaction in a laboratory. The reaction is a simple one: $A + B ightleftharpoons C + D$. The problem gives us a very specific and somewhat unusual starting condition. Instead of just starting with reactants, we are told that initially, **all four species**—both reactants and products—are present at a concentration of $1	ext{ M}$ each.

To keep track of how these concentrations change as the system moves towards equilibrium, we use a powerful tool called an **ICE table** (Initial, Change, Equilibrium).

We start by filling in the 'Initial' row with $1	ext{ M}$ for $A$, $B$, $C$, and $D$. Now, we need to determine the 'Change'. Let's assume the reaction proceeds in the forward direction to reach equilibrium. This means a certain amount of $A$ and $B$ will be consumed. Let's call this amount $x$. Consequently, the concentrations of $A$ and $B$ will decrease by $x$ ($-x$), while the concentrations of the products $C$ and $D$ will increase by the exact same amount $x$ ($+x$).

Adding the initial concentrations and the changes gives us our equilibrium concentrations:
*   $[A]_{eq} = 1 - x$
*   $[B]_{eq} = 1 - x$
*   $[C]_{eq} = 1 + x$
*   $[D]_{eq} = 1 + x$

### The Master Equation: Equilibrium Constant

The heart of any chemical equilibrium problem is the equilibrium constant expression. The equilibrium constant, $K_{eq}$, is defined as the ratio of the product of the equilibrium concentrations of the products to the product of the equilibrium concentrations of the reactants.

For our reaction, the expression is:
$$K_{eq} = \frac{[C][D]}{[A][B]}$$

We are given that $K_{eq} = 100$ at $298	ext{ K}$. Now, we simply substitute the equilibrium expressions we derived from our ICE table into this master equation:
$$100 = \frac{(1+x)(1+x)}{(1-x)(1-x)}$$

### The Algebraic Elegance

At first glance, this equation might look like it's going to turn into a messy quadratic equation. If you were to expand the numerator and the denominator, you would indeed get a quadratic. However, take a closer look at the structure of the equation!

The numerator is $(1+x)$ multiplied by itself, which is $(1+x)^2$. The denominator is $(1-x)$ multiplied by itself, which is $(1-x)^2$. This means the entire right side of the equation is a perfect square:
$$100 = \left(\frac{1+x}{1-x}ight)^2$$

This is where we can use a little algebraic elegance to save a lot of time. By taking the square root of both sides, we completely bypass the need for the quadratic formula. The square root of $100$ is $10$, so our equation simplifies beautifully to:
$$10 = \frac{1+x}{1-x}$$

Now, it's just a matter of simple linear algebra. We cross-multiply to get:
$$10(1 - x) = 1 + x$$

Expanding the bracket gives:
$$10 - 10x = 1 + x$$

Grouping the $x$ terms on one side and the constants on the other, we find:
$$9 = 11x$$

Solving for $x$, we get:
$$x = \frac{9}{11} \approx 0.818$$

### The Final Piece of the Puzzle

We have found the value of $x$, but we are not quite done yet. The question specifically asks for the equilibrium concentration of $D$.

Let's look back at our trusty ICE table. The equilibrium concentration of $D$ is given by the expression $1 + x$. Now that we know $x = 0.818$, we simply substitute this value back in:
$$[D]_{eq} = 1 + 0.818 = 1.818	ext{ M}$$

And there we have it! By carefully setting up our initial conditions, applying the law of mass action, and using a clever algebraic shortcut, we've arrived at the final answer. The equilibrium concentration of $D$ is **$1.818	ext{ M}$**.

Animated Solution for Chemistry - Chemical Equilibrium: The equilibrium constant at $298 K$ for a reaction, $A + B ⇌ C + D$ is $100$ . If the initial concentrations of all the four species were $1 M$ each, then equilibrium concentration of $D$ (in $mol L^{- 1}$ ) will be

Select Answer:

Visualized Solution

\text{Initial State}

\text{ICE Table Setup}

\text{Equilibrium Constant Expression}

\text{Simplifying the Equation}

\text{Solving for } x

\text{Final Concentration of D}

The Sigma Insight: Law of Mass Action

Setting the Stage

The Master Equation

The Algebraic Elegance

The Final Piece of the Puzzle

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