Problem 111 - Chemistry

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 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.

Step-by-Step Solution

Answer

The equilibrium concentrations of the species in the reaction are: \([\mathrm{NOCl}] = 0.0032 \, \text{M}\), \([\mathrm{NO}] = 1.9968 \, \text{M}\), and \([\mathrm{Cl}_{2}] = 0.9984 \, \text{M}\).

Step 1: Write the equilibrium expression

For the given reaction: $$2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g)$$ The equilibrium expression, K, is: $$K = \frac{[\mathrm{NO}]^2 [\mathrm{Cl}_{2}]}{[\mathrm{NOCl}]^2}$$ Where K is given as \(1.6 \times 10^{-5}\) at \(35^{\circ}\mathrm{C}\).

Step 2: Set up the ICE table

We are given the initial amounts (in moles) of NO and Cl₂ and the fixed volume of the flask. Therefore, we can calculate the initial concentrations: $$[\mathrm{NO}]_{0} = \frac{2.0 \, \text{moles}}{1.0 \, \text{L}} = 2.0 \, \text{M}$$ $$[\mathrm{Cl}_{2}]_{0} = \frac{1.0 \, \text{moles}}{1.0 \, \text{L}} = 1.0 \, \text{M}$$ $$[\mathrm{NOCl}]_{0} = 0 \, \text{M}$$ Now, set up the ICE table: | | NOCl | NO | Cl₂ | |------|------|----|-----| | I | 0 | 2.0| 1.0 | | C | +2x | -2x| -x | | E | 2x | 2.0-2x| 1.0-x | Notice that the stoichiometry of the reaction shows equal moles of NOCl converting into NO, so the change in concentrations will follow a ratio of 2:2, or 1:1.

Step 3: Solve for equilibrium concentrations using ICE table and K

Using K and the ICE table values, we have: \(K = \frac{[\mathrm{NO}]^2 [\mathrm{Cl}_{2}]}{[\mathrm{NOCl}]^2} = \frac{(2.0-2x)^2(1.0-x)}{(2x)^2}\) Plug the given value of K: \(1.6 \times 10^{-5} = \frac{(2.0-2x)^2(1.0-x)}{(2x)^2}\) To solve for x, quadratic equation solvers can be used. The value of x will give the changes in the concentrations of the reactants and products at equilibrium. Upon solving, we get: \(x = 0.0016\) Now, plug the value of x into the E row of the ICE table: $$[\mathrm{NOCl}] = 2x = 2(0.0016) = 0.0032 \, \text{M}$$ $$[\mathrm{NO}] = 2.0 - 2x = 2.0 - 2(0.0016) = 1.9968 \, \text{M}$$ $$[\mathrm{Cl}_{2}] = 1.0 - x = 1.0 - 0.0016 = 0.9984 \, \text{M}$$ So, the equilibrium concentrations are: $$[\mathrm{NOCl}] = 0.0032 \, \text{M}$$ $$[\mathrm{NO}] = 1.9968 \, \text{M}$$ $$[\mathrm{Cl}_{2}] = 0.9984 \, \text{M}$$

Key Concepts

ICE Table

An ICE table is a helpful tool used to handle problems in chemical equilibrium. It helps to determine the concentrations of reactants and products at equilibrium. "ICE" stands for:

• Initial: the starting concentrations of reactants and products.
• Change: the changes in concentrations as the system moves towards equilibrium.
• Equilibrium: the final concentrations once the system has reached equilibrium.

In the given reaction, we set up the ICE table with initial concentrations:

• 2.0 M for NO
• 1.0 M for Cl₂
• 0 M for NOCl

The changes (C) are determined by stoichiometry, where NO and NOCl change by -2x and +2x, respectively, and Cl₂ by -x. Ultimately, this gives the equilibrium row (E) in terms of x. The ICE table simplifies complex calculations by structuring the changes and aiding in applying the equilibrium constant equation.

Equilibrium Constant (K)

The equilibrium constant (K) is a critical parameter in the study of chemical equilibrium. It quantifies the ratio of the concentrations of products to reactants when a reaction is at equilibrium, raised to the power of their coefficients in the balanced equation.
For the equation: \(2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g) + \mathrm{Cl}_{2}(g)\),
the equilibrium expression is:
\[ K = \frac{[\mathrm{NO}]^2 [\mathrm{Cl}_{2}]}{[\mathrm{NOCl}]^2} \]
K helps you predict the direction in which a reaction will proceed to reach equilibrium. In our scenario, K equals \(1.6 \times 10^{-5}\), a small number indicating that at equilibrium, products are only present in low concentrations compared to reactants. Calculating the exact equilibrium concentrations involves substituting known values and solving for unknowns to achieve this K value in the equilibrium expression.

Molar Concentration

Molar concentration, also known as molarity, is a measure of the concentration of a solute in a solution. It describes how many moles of solute are present in one liter of solution, expressed in moles per liter (M).
In this context, the initial molar concentrations of the reactants were provided:

• \([\mathrm{NO}] = 2.0 \, \text{M}\)
• \([\mathrm{Cl}_{2}] = 1.0 \, \text{M}\)

The molar concentration of NOCl was initially zero, as it is used as the product from the reaction.
As the reaction progresses towards equilibrium, these concentrations change and are calculated using the ICE table. Knowing the changes (which are functions of x), the equilibrium molar concentrations are deduced, providing a clear picture of the distribution of species when chemical equilibrium is achieved.