Question

For the formation of ammonia, the equilibrium constant is calculated to be \(5.2 \times 10^{-5}\) at \(25^{\circ} \mathrm{C}\) . After analysis, it is determined that \(\left[\mathrm{N}_{2}\right]=2.00 \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]=0.80 \mathrm{M} .\) How many grams of ammonia are in the 10. L reaction vessel at equilibrium? Use the following equilibrium equation. $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftarrows 2 \mathrm{NH}_{3}(g)$$

Short answer

1.241 g of \(\mathrm{NH}_3\) are in the 10 L reaction vessel at equilibrium.

Step-by-step solution

- Write the equilibrium expression

The equilibrium constant expression for the reaction is given by: \[ K_c = \frac{[\mathrm{NH}_3]^2}{[\mathrm{N}_2][\mathrm{H}_2]^3} \] Given that \( K_c = 5.2 \times 10^{-5} \), substitute this value in the expression and solve for \( [\mathrm{NH}_3] \).

- Substitute known values into the equilibrium expression

Given: \( [\mathrm{N}_2] = 2.00 \; M \) and \( [\mathrm{H}_2] = 0.80 \; M \).Substitute these values into the Kc expression: \[ 5.2 \times 10^{-5} = \frac{[\mathrm{NH}_3]^2}{(2.00)(0.80)^3} \]

- Simplify the equation

First calculate the denominator:\[ (2.00)(0.80)^3 = 2.00 \times 0.512 = 1.024 \]So the new equation is:\[ 5.2 \times 10^{-5} = \frac{[\mathrm{NH}_3]^2}{1.024} \]

- Solve for \( [\mathrm{NH}_3] \)

Rearrange the equation to solve for \( [\mathrm{NH}_3] \): \[ [\mathrm{NH}_3]^2 = (5.2 \times 10^{-5}) \times 1.024 = 5.3248 \times 10^{-5} \]Take the square root of both sides: \[ [\mathrm{NH}_3] = \sqrt{5.3248 \times 10^{-5}} \approx 7.30 \times 10^{-3} \; M \]

- Convert concentration to moles

We need to find the number of moles of ammonia in the 10 L vessel. If \( [\mathrm{NH}_3] = 7.30 \times 10^{-3} \; M \), then:\[ \text{moles of } \mathrm{NH}_3 = (7.30 \times 10^{-3}) \; M \times 10 \; L = 7.30 \times 10^{-2} \; \text{moles} \]

- Convert moles to grams

Using the molar mass of ammonia (\(\mathrm{NH}_3\)), which is approximately \( 17 \; g/mol \), we can convert moles to grams: \[ \text{grams of } \mathrm{NH}_3 = 7.30 \times 10^{-2} \; \text{moles} \times 17 \; g/mol = 1.241 \; g \]

Key concepts

chemical equilibrium

Chemical equilibrium occurs in a reversible reaction when the rate of the forward reaction equals the rate of the backward reaction. At this point, the concentrations of reactants and products remain constant over time. This does not mean that the reactions stop; rather, they continue to happen, but at the same rate, so there is no net change in the concentration of chemicals involved.

The equilibrium constant, represented as \( K_c \), is a numerical value that describes the ratio of the concentrations of products to the concentrations of reactants at equilibrium. For the given exercise, the equilibrium constant for the formation of ammonia is calculated as \( 5.2 \times 10^{-5} \) at \( 25^{\textdegree} \text{C} \).

This value informs us about the extent of the reaction: a small \( K_c \) indicates the equilibrium favors reactants, whereas a large \( K_c \) indicates it favors products.

molar concentration

Molar concentration, often called molarity, is a way of expressing the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution (mol/L or M).

In the given exercise, we are provided with the molar concentrations of nitrogen \( ( \text{N}_2 ) \) and hydrogen \( ( \text{H}_2 ) \):

• \( [ \text{N}_2 ] = 2.00 \text{ M} \)
• \( [ \text{H}_2 ] = 0.80 \text{ M} \)

These values are essential as we use them to determine the molarity of ammonia \( ( \text{NH}_3 ) \) at equilibrium by substituting them into the equilibrium constant expression.

reaction vessels

A reaction vessel is a container where chemical reactions take place. Common reaction vessels include test tubes, flasks, and beakers. The size and type of the vessel can significantly affect the reaction by influencing factors like volume, pressure, and temperature.

In the exercise, a 10 L reaction vessel is used to find out how many grams of ammonia are present at equilibrium. The volume of the reaction vessel allows us to convert the molar concentration of ammonia to moles, and then to grams. Knowing the volume (10 L) of the reaction vessel is crucial for performing these conversions correctly.

stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It is based on the balanced chemical equation, which provides the ratio of the molecules or moles involved.

In this exercise, the balanced equation for the formation of ammonia is:
\[ \text{N}_2 (g) + 3 \text{H}_2 (g) \rightleftarrows 2 \text{NH}_3 (g) \]
This tells us that one mole of nitrogen reacts with three moles of hydrogen to produce two moles of ammonia. Using stoichiometric principles, we can plug the known concentrations into the equilibrium expression and solve for the unknown concentration of ammonia \( [ \text{NH}_3 ] \). Following this, we can further use stoichiometry to convert this concentration into the number of grams of ammonia in the reaction vessel.