For the reaction, A + B 2C. The value of equilibrium constant is 100 at 298 K. If the initial co

Q: Why did we calculate $Q_c$ first?

To determine the direction of the reaction. If $Q_c > K_c$, the reaction would go backward, and the change would be $+x, +x, -2x$. Since $Q_c < K_c$, we confirmed it goes forward.

Q: Why is the change for C $+2x$ and not $+x$?

The stoichiometric coefficient of C in the balanced equation is 2. For every 1 mole of A that reacts, 2 moles of C are produced.

Q: Can we take the negative square root of 100?

Mathematically, yes (-10). But physically, it would lead to an $x$ value that makes equilibrium concentrations negative, which is impossible in reality.

Q: What if the equation wasn't a perfect square?

You would have to cross-multiply and solve the resulting quadratic equation using the standard quadratic formula.

Q: Why does the unit of concentration not affect $K_c$ here?

Because the number of moles of gaseous products equals the number of moles of gaseous reactants ($\Delta n_g = 0$), the units cancel out completely.

Question

For the reaction, $A + B ightleftharpoons 2C$
The value of equilibrium constant is 100 at 298 K. If the initial concentration of all the three species is 1 M each, then the equilibrium concentration of C is $x 	imes 10^{-1}$ M. The value of $x$ is ...... .
(Nearest integer)

Sigma Percentile · Accepted Answer

### The Art of the ICE Table: Mastering Chemical Equilibrium

Chemical equilibrium is a delicate dance of molecules, where the forward and reverse reactions occur at the exact same rate. To decode this dance, chemists rely on a powerful tool: the **ICE Table** (Initial, Change, Equilibrium). Let's dive into a classic problem that perfectly illustrates its elegance.

### Analyzing the Setup

We are given the reaction:

$$A + B ightleftharpoons 2C$$

Initially, we have $1	ext{ M}$ of each species. Before we jump into calculating changes, we must ask a crucial question: *Which way will the reaction shift to reach equilibrium?*

To answer this, we calculate the reaction quotient, $Q_c$:

$$Q_c = \frac{[C]_0^2}{[A]_0[B]_0} = \frac{1^2}{1 	imes 1} = 1$$

Since $Q_c (1)$ is less than the equilibrium constant $K_c (100)$, the reaction must proceed in the **forward direction** to produce more $C$ and consume $A$ and $B$.

### Building the ICE Table

Knowing the direction, we can confidently define our changes. Let $x$ be the amount of $A$ that reacts. Because of the $1:1:2$ stoichiometry, $B$ will also decrease by $x$, and $C$ will increase by $2x$.

Our equilibrium concentrations become:
- $[A] = 1 - x$
- $[B] = 1 - x$
- $[C] = 1 + 2x$

### The Master Equation and a Mathematical Trick

Now, we substitute these expressions into the equilibrium constant formula:

$$K_c = \frac{[C]^2}{[A][B]}$$

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

Here is where many students fall into a trap. Expanding the squares leads to a messy quadratic equation. But look closely! Both the numerator and the denominator are perfect squares, and so is $100$. We can bypass the quadratic nightmare by simply taking the **square root of both sides**:

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

$$10 = \frac{1+2x}{1-x}$$

### Final Calculation

With a simple linear equation, solving for $x$ is a breeze:

$$10(1-x) = 1+2x$$

$$10 - 10x = 1 + 2x$$

$$12x = 9 \Rightarrow x = 0.75$$

Finally, we find the equilibrium concentration of $C$:

$$[C] = 1 + 2x = 1 + 2(0.75) = 2.5	ext{ M}$$

The question asks for the answer in the format $x 	imes 10^{-1}$. We can rewrite $2.5$ as $25 	imes 10^{-1}$. Thus, our final integer answer is **$25$**.

Enter Numerical Value:

Visualized Solution

\text{Initial State}

\text{Reaction Direction}

\text{Defining the Change}

\text{Equilibrium Concentrations}

\text{Equilibrium Constant Expression}

\text{Substituting Values}

\text{Taking Square Root}

\text{Solving for } x

\text{Finding } [C]

\text{Formatting the Answer}

The Sigma Insight: Law of Mass Action

The Art of the ICE Table

Analyzing the Setup

Building the ICE Table

The Master Equation and a Mathematical Trick

Final Calculation

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