Question

A mixture of \(0.10 \mathrm{~mol}\) of \(\mathrm{NO}, 0.050 \mathrm{~mol}\) of \(\mathrm{H}_{2}\), and \(0.10 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}\) is placed in a 1.0-L vessel at \(300 \mathrm{~K}\). The following equilibrium is established: $$ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ At equilibrium \([\mathrm{NO}]=0.062 M .\) (a) Calculate the equilibrium concentrations of \(\mathrm{H}_{2}, \mathrm{~N}_{2}\), and \(\mathrm{H}_{2} \mathrm{O}\). (b) Calculate \(K_{\mathrm{c}}\).

Short answer

(a) Equilibrium concentrations: \[H_2=0.012 \, M,\] \[N_2=0.019 \, M,\] \[H_2O=0.138 \, M\] (b) \[K_c = 56.6\]

Step-by-step solution

Balanced chemical equation

The balanced chemical equation is already provided: \[2 NO(g) + 2 H_2(g) \rightleftharpoons N_2(g) + 2 H_2O(g)\] Step 2: Set up the ICE table

ICE table

Set up an ICE table to organize the initial concentrations, the changes in concentrations during the reaction, and the equilibrium concentrations of all species: $$ \begin{array}{c|cccc} & 2NO & +2H_2 & \rightleftharpoons & N_2 & +2H_2O \\ \hline Initial & 0.10 \, M & 0.050\, M & & 0 & 0.10 \, M \\ Change & -2x & -2x & & +x & +2x \\ Equilibrium & 0.10-2x & 0.050-2x & & x & 0.10+2x \end{array} $$ Step 3: Equilibrium concentrations

Equilibrium concentrations

We are given that the equilibrium concentration of NO is 0.062 M. This value can be used to determine the concentrations of other species. \[0.10-2x=0.062\] \[x = \frac{0.10 - 0.062}{2}= 0.019\] Now, we can determine the equilibrium concentrations of the other species: \[H_2: 0.050-2x = 0.050 - 2(0.019) = 0.012\, M\] \[N_2: x = 0.019\, M\] \[H_2O: 0.10+2x = 0.10 + 2(0.019) = 0.138\, M\] Step 4: Calculate Kc

Calculate Kc

Finally, use the equilibrium concentrations to calculate the equilibrium constant Kc: \[K_c = \frac{[N_2][H_2O]^2}{[NO]^2[H_2]^2} = \frac{(0.019)(0.138)^2}{(0.062)^2(0.012)^2} = 56.6\] #Answer#: (a) Equilibrium concentrations: \[[H_2]=0.012 \, M,\] \[[N_2]=0.019 \, M,\] \[[H_2O]=0.138 \, M\] (b) \[K_c = 56.6\]

Key concepts

ICE Table

The ICE table is an essential tool in chemical equilibrium calculations, representing the Initial concentrations, Changes in concentration, and Equilibrium concentrations of the reactants and products involved in a chemical process. To create an ICE table:

• Start by writing the balanced chemical equation.
• List each species participating in the equation with their initial concentrations.
• Identify the change required as the reaction progresses toward equilibrium, using variables like 'x' to denote unknown changes.
• Finally, express equilibrium concentrations as formulas based on initial amounts and changes.

This table helps visualize how concentrations vary during a reaction and simplifies the calculations needed to determine equilibrium states. In our exercise, knowing that the initial concentration of \(NO\) is \(0.10 \, M\) and considering a decrease represented by \(2x\), helps track what happens to \(H_2\) and form \(N_2\) and \(H_2O\).

Equilibrium Constant (Kc)

The equilibrium constant, denoted as \(K_c\), reflects the ratio of concentrations of products to reactants, each raised to the power of their coefficients in the balanced equation, at equilibrium. It's given by:\[K_c = \frac{[products]}{[reactants]}\]The value of \(K_c\) is specific to a particular reaction at a given temperature and indicates the extent to which a reaction proceeds. A large \(K_c\) suggests that the equilibrium heavily favors products, whereas a small \(K_c\) implies that reactants predominate. In our solved exercise, after substituting the equilibrium concentrations into the equilibrium expression, we find \(K_c = 56.6\), indicating significant product formation under the given conditions.

Using \(K_c\), we can predict how changes in conditions will affect the equilibrium position according to Le Chatelier's principle, making it a crucial concept for understanding chemical systems.

Equilibrium Concentrations

Equilibrium concentrations refer to the amounts of reactants and products present once a chemical reaction reaches a state where the rate of the forward reaction equals the rate of the backward reaction. To find these concentrations:

• Use known equilibrium data, such as one established concentration, to solve for unknowns.
• Apply principles from the ICE table, using arithmetic to find the shift in concentrations.

In the given problem, \[H_2\] is calculated after determining \x\ from the change in \[NO\]. This allows calculation for \[N_2\] and \[H_2O\] using their respective formulas from the ICE table.

Understanding equilibrium concentrations assists in manipulating reactions and optimizing conditions in industrial or lab settings, where specific concentrations of products may be desired.

Reaction Quotient

The reaction quotient, \(Q_c\), is similar to the equilibrium constant but is calculated using the concentrations at any point in time during a reaction, not just at equilibrium. It is given by the same expression as \(K_c\), but involves the initial concentrations instead. The reaction quotient is used to predict the direction in which a reaction will proceed:

• If \(Q_c > K_c\), the reaction will shift towards the reactants to reach equilibrium.
• If \(Q_c < K_c\), the reaction will shift towards the products.
• If \(Q_c = K_c\), the system is already at equilibrium.

Though \(Q_c\) was not explicitly calculated in the exercise, understanding its role provides insight into anticipating changes in a reaction when concentrations or conditions are altered. It's a vital tool for chemists in diagnosing how far a system is from achieving equilibrium.