Question

At a particular temperature a \(2.00-\mathrm{L}\) flask at equilibrium contains \(2.80 \times 10^{-4}\) mole of \(\mathrm{N}_{2}, 2.50 \times 10^{-5}\) mole of \(\mathrm{O}_{2}\), and \(2.00 \times 10^{-2}\) mole of \(\mathrm{N}_{2} \mathrm{O}\). Calculate \(K\) at this temperature for the reaction $$ 2 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{~N}_{2} \mathrm{O}(g) $$ If \(\left[\mathrm{N}_{2}\right]=2.00 \times 10^{-4} M,\left[\mathrm{~N}_{2} \mathrm{O}\right]=0.200 M\), and \(\left[\underline{ \left.\mathrm{O}_{2}\right]}=\right.\) \(0.00245 M\), does this represent a system at equilibrium?

Short answer

The equilibrium constant (K) for the reaction is 64. The reaction quotient (Q) calculated using the given concentrations is approximately 4039. Since Q > K, the given system is not at equilibrium and will shift to consume the products (N2O) and form the reactants (N2 and O2) to reach equilibrium.

Step-by-step solution

Calculate molar concentrations at equilibrium

To find the molar concentration of each substance, we'll divide the number of moles by the volume of the flask (2.00 L). \[ [N_2]_e = \frac{2.80 \times 10^{-4} \,\text{mol}}{2.00 \,\text{L}} = 1.40 \times 10^{-4}\, M \] \[ [O_2]_e = \frac{2.50 \times 10^{-5} \,\text{mol}}{2.00 \,\text{L}} = 1.25 \times 10^{-5}\, M \] \[ [N_2O]_e = \frac{2.00 \times 10^{-2} \,\text{mol}}{2.00 \,\text{L}} = 1.00 \times 10^{-2}\, M \]

Calculate the equilibrium constant (K)

Using the molar concentrations of each substance at equilibrium, we can now substitute these values into the equilibrium expression to calculate K. \[ K_c = \frac{[N_2O]^2}{[N_2]^2 \times [O_2]} = \frac{(1.00 \times 10^{-2})^2}{(1.40 \times 10^{-4})^2 \times (1.25 \times 10^{-5})} \] \[ K_c = 64 \]

Calculate the reaction quotient (Q)

To check if the given concentrations represent a system at equilibrium, we'll calculate the reaction quotient (Q) using the given concentrations: \[ Q_c = \frac{[N_2O]^2}{[N_2]^2 \times [O_2]} = \frac{(0.200)^2}{(2.00 \times 10^{-4})^2 \times (0.00245)} \] \[ Q_c \approx 4039 \]

Compare K and Q

Now that we have calculated the equilibrium constant (K) and the reaction quotient (Q), we'll compare them to determine if the given concentrations represent a system at equilibrium. Since Q > K (4039 > 64), the system is not at equilibrium. The reaction will need to shift in the direction that consumes the products (N2O) and forms the reactants (N2 and O2) to reach equilibrium.

Key concepts

Equilibrium Constant (K)

The equilibrium constant, denoted as K, is a fundamental concept in chemistry that represents the ratio of the concentration of products to the concentration of reactants for a reversible reaction at equilibrium. The concentration of each reactant and product is raised to the power of its coefficient in the balanced chemical equation.

For the equilibrium reaction:
\[2 \textrm{N}_2(g) + \textrm{O}_2(g) \rightleftharpoons 2 \textrm{N}_2\textrm{O}(g),\]
the equilibrium constant expression would be:
\[K_c = \frac{[N_2O]^2}{[N_2]^2 \times [O_2]}\]
where K_c is the equilibrium constant for concentrations and the square brackets denote the molar concentration of each species. The subscript 'c' signifies that the equilibrium constant is expressed in terms of concentrations. It is crucial to note that K is temperature-dependent and constant for a given reaction at a fixed temperature.

The equilibrium constant provides insight into the extent of a reaction; a large K indicates a mix with a high concentration of products relative to reactants, typically suggesting that the reaction strongly favors the formation of products. Conversely, a small K implies that the reaction yields little products and will favor the reactants.

Reaction Quotient (Q)

The reaction quotient, Q, is a measure similar to the equilibrium constant but describes the ratio of molar concentrations of products to reactants at any moment in time, not necessarily at equilibrium. The expression for Q is identical to that of K, but it uses the current concentrations instead of those at equilibrium. Thus, for our given reaction:
\[Q_c = \frac{[N_2O]^2}{[N_2]^2 \times [O_2]}\]

The value of Q provides important information about the system's status. If Q equals K, the system is at equilibrium. If Q is greater than K, the reaction will proceed in the reverse direction to reach equilibrium, consuming products and generating reactants. Conversely, if Q is less than K, the reaction tends to go forward, creating more products and using up reactants until equilibrium is established.

In our exercise, since Q is significantly larger than K, the system is not in equilibrium and will shift left, turning products (\textrm{N}_2\textrm{O}) into reactants (\textrm{N}_2 and \textrm{O}_2) in order to reach equilibrium.

Molar Concentration

Molar concentration (also called molarity) of a solution is the amount of a solute that is present per unit volume of the solution. It is a measure of the concentration of a substance within a given volume of liquid and is expressed as moles per liter (M). The formula for calculating molar concentration is:
\[ \text{Molar Concentration} = \frac{\text{Number of Moles of Solute}}{\text{Volume of Solution in Liters}} \]

Molar concentration allows chemists to quickly understand the amount of substance involved in a chemical reaction. In the step-by-step solution, molar concentrations of \textrm{N}_2, \textrm{O}_2, and \textrm{N}_2\textrm{O} were calculated by dividing the number of moles of each substance by the volume of the flask that contains them.

Understanding how to calculate and apply molar concentration is critical when discussing chemical equilibrium, as the equilibrium constant and reaction quotient rely on these concentrations. Molarity is especially useful because it directly relates to the volume of the solution, which makes experimental tasks such as diluting or mixing solutions accurately much more straightforward for scientists and students alike.