Question

Ammonia, a very important industrial chemical, is produced by the direct combination of the following elements under carefully controlled conditions: $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ Suppose, in an experiment, that the reaction mixture is analyzed after equilibrium is reached, and it is found, at a particular temperature, that \(\left[\mathrm{NH}_{3}\right]=0.25 \mathrm{M}\), \(\left[\mathrm{H}_{2}\right]=3.4 \times 10^{-3} M,\) and \(\left[\mathrm{N}_{2}\right]=5.3 \times 10^{-5} M .\) Calculate the value of \(K\) at this temperature.

Short answer

The equilibrium constant, K, for the reaction at this temperature can be calculated using the given concentrations of NH3, H2, and N2, and the expression \(K = \frac{[NH_3]^2}{[N_2] \times [H_2]^3}\). By plugging in the given values and solving, we find that \(K \approx 3.125 \times 10^{11}\).

Step-by-step solution

Understand the relationship between equilibrium constant K and concentrations

The equilibrium constant, K, is the ratio of the concentration of products to the concentration of reactants, with each concentration raised to the power of its stoichiometric coefficient. For the given reaction, K is calculated as follows: \(K = \frac{[NH_3]^2}{[N_2] \times [H_2]^3}\)

Plug in the given equilibrium concentrations

We are given that \([NH_{3}] = 0.25M\), \([H_{2}] = 3.4 \times 10^{-3}M\), and \([N_{2}] = 5.3 \times 10^{-5}M\). Substitute these values into the expression for K above: \(K = \frac{(0.25)^2}{(5.3 \times 10^{-5})(3.4 \times 10^{-3})^3}\)

Calculate the value of K

Now, calculate K by following the order of operations (parentheses, exponents, multiplication, and division): \(K = \frac{(0.25)^2}{(5.3 \times 10^{-5})(3.4 \times 10^{-3})^3} = \frac{0.0625}{(5.3 \times 10^{-5})(3.9 \times 10^{-8})} = \frac{0.0625}{2 \times 10^{-12}}\) \(K = 3.125 \times 10^{11}\) The equilibrium constant, K, for the reaction at this temperature is approximately \(3.125 \times 10^{11}\).

Key concepts

Equilibrium Constant (K)

When we talk about the equilibrium constant, denoted as "K", we're talking about a special number that represents the ratio of product concentrations to reactant concentrations in a chemical reaction at equilibrium. This ratio is tweaked by the stoichiometric coefficients given in the balanced chemical equation. The formula for the equilibrium constant is derived directly from this relationship.
For the reaction of nitrogen gas with hydrogen gas to form ammonia (\(\mathrm{N}_{2}(g) + 3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\)), K is expressed as:

• \( K = \frac{[\mathrm{NH}_3]^2}{[\mathrm{N}_2] \times [\mathrm{H}_2]^3} \)

The exponents in the expression come from the coefficients in the balanced equation. These coefficients indicate how many molecules of each substance are involved in the reaction. Understanding K is crucial because it tells us how the concentrations change as the system reaches equilibrium, giving insight into the position of the equilibrium favoring products or reactants at a certain temperature.

Stoichiometry

Stoichiometry is like the recipe of a chemical reaction. It tells us the proportions in which reactants combine and products form. In our ammonia production reaction, the coefficients "1" for \( \mathrm{N}_{2}(g) \), "3" for \( \mathrm{H}_{2}(g) \), and "2" for \( \mathrm{NH}_{3}(g) \) guide us in understanding the quantities needed for the reaction.
These coefficients are essential when calculating the equilibrium constant, as they determine the exponents used in the K expression. You raise the concentration of each substance to the power of its stoichiometric coefficient:

• \([\mathrm{NH}_3]^2\) because of the "2" in front of \(\mathrm{NH}_3\).
• \([\mathrm{H}_2]^3\) because of the "3" in front of \(\mathrm{H}_2\).

This aspect highlights the importance of stoichiometry not only in balancing equations but also in determining reaction dynamics at equilibrium. Every small number in a balanced equation can have a big impact when computing something as significant as the equilibrium constant.

Concentration Calculations

Calculating concentrations at equilibrium is a fundamental skill in chemistry, especially when it comes to applying them in the context of the equilibrium constant. Given the concentrations after the reaction has reached equilibrium, we plug these values into the K expression.
In our example, we're provided with:

• \([\mathrm{NH}_3] = 0.25\, \mathrm{M}\)
• \([\mathrm{H}_2] = 3.4 \times 10^{-3}\, \mathrm{M}\)
• \([\mathrm{N}_2] = 5.3 \times 10^{-5}\, \mathrm{M}\)

Substitute these into \( K = \frac{[\mathrm{NH}_3]^2}{[\mathrm{N}_2] \times [\mathrm{H}_2]^3} \), like solving a puzzle. It follows the step-by-step mechanism of the equation: calculate each part separately, then divide the numerator by the product of the reactants raised to their respective powers.
After doing the math, this helps us get a clearer picture of the state of equilibrium and the extent to which reactants are converted into products. It's the practical application of chemical concepts, transforming numbers into understanding of the reaction dynamics.