Animated Solution for Chemistry - Chemical Equilibrium: For the reaction equilibrium, ${\text{N}}_2{\text{O}}_4(g)\rightleftharpoons 2{\text{NO}}_2(g)$ the concentrations of ${\text{N}}_2{\text{O}}_4$ and ${\text{NO}}_2$ at equilibrium are $4.8\times 10^{-2}$ and $1.2\times 10^{-2}{\text{ mol L}}^{-1}$, respectively. The value of $K_c$ for the reaction is

Imagine a closed flask where a fascinating dance of molecules is taking place. On one side, we have the colorless dinitrogen tetroxide gas, ${\text{N}}_2{\text{O}}_4$. On the other side, we have the reddish-brown nitrogen dioxide gas, ${\text{NO}}_2$. These two gases are constantly converting into each other.

When the rate at which ${\text{N}}_2{\text{O}}_4$ breaks down into ${\text{NO}}_2$ perfectly matches the rate at which ${\text{NO}}_2$ molecules recombine to form ${\text{N}}_2{\text{O}}_4$, the system reaches a state of chemical equilibrium. At this point, the concentrations of both gases remain constant over time, even though the reactions are still actively happening at the microscopic level.

To quantify this state of equilibrium, chemists use a powerful tool called the Law of Mass Action. This law states that for a reversible reaction at equilibrium and a constant temperature, a certain ratio of reactant and product concentrations has a constant value, known as the equilibrium constant, $K_c$.

For our specific reaction:

The equilibrium constant expression is written by taking the concentration of the products and dividing it by the concentration of the reactants. Crucially, each concentration must be raised to the power of its stoichiometric coefficient from the balanced chemical equation.

Following the rule, the expression for $K_c$ becomes:

Notice the squared term for ${\text{NO}}_2$. This is a common place where students make silly mistakes. Because there is a '2' in front of ${\text{NO}}_2$ in the balanced equation, its concentration must be squared in the equilibrium expression.

The problem provides us with the exact concentrations of both gases at equilibrium: - $[{\text{N}}_2{\text{O}}_4]=4.8\times 10^{-2}{\text{ mol L}}^{-1}$ - $[{\text{NO}}_2]=1.2\times 10^{-2}{\text{ mol L}}^{-1}$

Now, we carefully substitute these values into our master equation:

Let's break down the calculation to avoid any errors. First, we square the numerator. Squaring $1.2$ gives us $1.44$, and squaring $10^{-2}$ gives us $10^{-4}$.

Next, we separate the decimal numbers from the powers of 10 to make the division easier:

Dividing $1.44$ by $4.8$ gives exactly $0.3$. For the powers of 10, dividing $10^{-4}$ by $10^{-2}$ leaves us with $10^{-2}$.

To express our final answer in standard scientific notation, we shift the decimal point one place to the right, which decreases the exponent by one:

This elegant result tells us the exact ratio of products to reactants at equilibrium for this specific temperature. Understanding how to manipulate these expressions and carefully track your units and exponents is a fundamental skill in mastering chemical equilibrium.