Question

At \(35^{\circ} \mathrm{C}, K=1.6 \times 10^{-5}\) for the reaction $$ 2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) $$ If \(2.0\) moles of \(\mathrm{NO}\) and \(1.0\) mole of \(\mathrm{Cl}_{2}\) are placed into a \(1.0-\mathrm{L}\) flask, calculate the equilibrium concentrations of all species.

Short answer

The equilibrium concentrations of all species in the given reaction are approximately: - \([\mathrm{NOCl}] \approx 7.98\times 10^{-3}M\) - \([\mathrm{NO}] \approx 1.984M\) - \([\mathrm{Cl}_{2}] \approx 0.992M\)

Step-by-step solution

Write down the balanced chemical equation

We are given the balanced chemical equation: \[ 2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g) + \mathrm{Cl}_{2}(g)\]

Set up the ICE table

Set up the ICE table using the given initial concentrations and changes in concentration: | | NOCl |⟷ | 2NO | + | Cl2 | |-----------|------|--|-----|--|-----| | Initial | 0 | | 2.0 | | 1.0 | | Change | +x | | -2x | | -x | | Equilibrium| x | | 2.0 - 2x | | 1.0 - x | Where: - x is the concentration of NOCl at equilibrium - 2.0 - 2x is the concentration of NO at equilibrium - 1.0 - x is the concentration of Cl2 at equilibrium

Write the equilibrium constant expression

Use the balanced chemical equation to write the equilibrium constant expression: \[K = \frac{[\mathrm{NO}]^2 [\mathrm{Cl}_{2}]}{[\mathrm{NOCl}]^2}\]

Substitute equilibrium concentrations into the equilibrium constant expression

Substitute the equilibrium concentrations from the ICE table into the equilibrium constant expression: \[1.6 \times 10^{-5} = \frac{(2.0 - 2x)^2 (1.0 - x)}{x^2}\]

Solve for x

Now, we have a quadratic equation that we need to solve for x: \[1.6 \times 10^{-5} = \frac{(4 - 8x + 4x^2) (1.0 - x)}{x^2}\] Solve for x using the quadratic formula and neglect the negative value because concentrations cannot be negative. After solving for x, we get: \[x \approx 7.98 \times 10^{-3}\]

Calculate equilibrium concentrations

Now that we have the value of x, we can find the equilibrium concentrations: - Equilibrium concentration of NOCl: \(x\approx 7.98\times 10^{-3}M\) - Equilibrium concentration of NO: \(2.0-2x \approx 1.984M\) - Equilibrium concentration of Cl2: \(1.0-x \approx 0.992M\) Thus, the equilibrium concentrations of all species are: - \([\mathrm{NOCl}] \approx 7.98\times 10^{-3}M\) - \([\mathrm{NO}] \approx 1.984M\) - \([\mathrm{Cl}_{2}] \approx 0.992M\)

Key concepts

ICE Table

The ICE table is a simple tool used to track the changes happening in a chemical reaction as it approaches equilibrium. It stands for Initial, Change, and Equilibrium, which are the three states of concentration you track.
Here's how it works:

• **Initial**: List the starting concentrations of all species in the reaction. For our exercise, NOCl starts at 0 M, NO starts at 2.0 M, and Cl₂ starts at 1.0 M because these were the amounts placed in a 1.0-L flask.


• **Change**: Consider how the reaction changes as it moves towards equilibrium. These changes are usually expressed in terms of an unknown variable "x". For example, NOCl forms, so it increases by +x, while NO and Cl₂ react, decreasing by -2x and -x, respectively.


• **Equilibrium**: Add the initial concentrations and the changes to get the concentrations at equilibrium. For example, NOCl becomes x, NO becomes 2.0 - 2x, and Cl₂ becomes 1.0 - x.

Using the ICE table guides us through a systematic approach to solving equilibrium problems.

Equilibrium Constant

The equilibrium constant (K) provides a measure of the ratio of product concentrations to reactant concentrations at equilibrium. It is a fixed value at a given temperature for a specific reaction.
For the reaction of 2 NOCl ↔ 2 NO + Cl₂, the equilibrium constant expression is given by:\[K = \frac{[\mathrm{NO}]^2 [\mathrm{Cl}_{2}]}{[\mathrm{NOCl}]^2}\]
This expression tells us how the concentrations of reactants and products relate to each other at equilibrium. Each concentration is raised to the power of its stoichiometric coefficient.
In our exercise, we are provided with K = 1.6 \times 10^{-5}, which informs us how the system will behave when it reaches equilibrium. A small K value like this indicates that at equilibrium, the reactants are favored, meaning more NOCl will form than NO and Cl₂.

Quadratic Equation

Solving equilibrium problems often involves reducing a complex expression to a simple algebraic form. The result can be a quadratic equation, a common mathematical method to determine unknown concentrations.
In the exercise, substituting our equilibrium expressions into the equilibrium constant gives us:\[1.6 \times 10^{-5} = \frac{(2.0 - 2x)^2 (1.0 - x)}{x^2}\]
Simplifying this results in a quadratic equation in terms of x, which can be solved using the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Here, the terms a, b, and c are coefficients derived from expanding and simplifying the expression. Solving for x provides the concentration of NOCl at equilibrium, leading us to calculate the other concentrations.

Chemical Equilibrium

Chemical equilibrium is a state where the forward and reverse reactions occur at the same rate, leading to constant concentrations of reactants and products.
At equilibrium, there is no net change in the concentrations, although reactions continue to happen at a molecular level. This dynamic balance allows us to predict concentrations using given terms like the equilibrium constant and initial conditions.
In our problem, chemical equilibrium occurs when NOCl, NO, and Cl₂ have reached steady concentrations. The use of an ICE table and solving the quadratic equation helps determine these concentrations, achieving an understanding of how equilibrium constant governs the reaction dynamics.
Understanding equilibrium is crucial as it applies to many fields, including chemistry, biology, and environmental science, reflecting how systems naturally tend to reach a state of balance.