Question

Ammonia, a very important industrial chemical, is produced by the direct combination of the 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 $[{\mathrm{NH}}_3(g)]=0.34M$ $[{\mathrm{H}}_2(g)]=2.1\times {10}^{-3}M,$ and $[{\mathrm{N}}_2(g)]=4.9\times {10}^{-4}M.$ Calculate the value of $K$ at this temperature.

Short answer

The value of K at this temperature, calculated using the provided equilibrium concentrations, is approximately $68.76$.

Step-by-step solution

Write the expression for K

For the given balanced chemical equation: $${\mathrm{N}}_2(g)+3{\mathrm{H}}_2(g)\rightleftharpoons 2{\mathrm{NH}}_3(g)$$ We can write the equilibrium constant expression as: $$K=\frac{[{\mathrm{NH}}_3{]}^2}{[{\mathrm{N}}_2][{\mathrm{H}}_2{]}^3}$$

Substitute the given concentrations

We are given the following concentrations: $$\begin{gathered} [{\mathrm{NH}}_3(g)]=0.34\text{M} \\ [{\mathrm{H}}_2(g)]=2.1\times {10}^{-3}\text{M} \\ [{\mathrm{N}}_2(g)]=4.9\times {10}^{-4}\text{M} \end{gathered}$$ Now, substitute these values into the K expression: $$K=\frac{(0.34{)}^2}{(4.9\times {10}^{-4})(2.1\times {10}^{-3}{)}^3}$$

Calculate K

Performing the calculation, we get: $$K=\frac{(0.34{)}^2}{(4.9\times {10}^{-4})(2.1\times {10}^{-3}{)}^3}=68.76$$ Therefore, the value of K at this temperature is approximately 68.76.

Key concepts

Equilibrium Constant

The equilibrium constant, denoted as $K$, is a vital concept in understanding chemical reactions at equilibrium. It provides insight into the ratio of the concentration of products to reactants at equilibrium for a given temperature. This constant is unique for each reaction and helps predict the direction of the reaction under different conditions.
For any reversible chemical reaction, the expression for $K$ can be derived from the balanced chemical equation. For example, in the reaction ${\mathrm{N}}_2(g)+3{\mathrm{H}}_2(g)\rightleftharpoons 2{\mathrm{NH}}_3(g)$, the equilibrium constant expression is: $$K=\frac{[{\mathrm{NH}}_3{]}^2}{[{\mathrm{N}}_2][{\mathrm{H}}_2{]}^3}$$ This formula reflects the concentrations of ammonia, nitrogen, and hydrogen at equilibrium. It's important to note that the exponents in the expression correspond to the coefficients in the balanced chemical equation.
Understanding $K$ allows chemists to:

• Predict whether a reaction will favor products or reactants at equilibrium.
• Determine the concentrations of different species in a mixture at equilibrium.
• Insight into reaction conditions required for optimal yield, especially in industrial applications.

Reaction Rates

Reaction rates are crucial in the study of chemical kinetics, describing how quickly a reaction proceeds. In the frame of chemical equilibrium, reaction rates reach a point where the forward and reverse reactions occur at equal speeds. This balance is what defines equilibrium itself.
When dealing with the equation ${\mathrm{N}}_2(g)+3{\mathrm{H}}_2(g)\rightleftharpoons 2{\mathrm{NH}}_3(g)$, both the production of ammonia and its dissociation occur simultaneously at equilibrium. The rate of the forward reaction (the combination of nitrogen and hydrogen to form ammonia) is equal to the rate of the reverse reaction (the breakdown of ammonia into nitrogen and hydrogen).
Factors affecting reaction rates include:

• Temperature: Increasing temperature generally increases reaction rates by providing more energy to the reacting molecules.
• Concentration: Higher concentrations of reactants can lead to increased collision rates, thus faster reactions.
• Catalysts: Catalysts speed up reactions without being consumed, often crucial in industrial processes.

Industrial Chemistry

Industrial chemistry applies chemical principles on a large scale to synthesize new products, improve processes, and enhance production efficiency. The synthesis of ammonia is a classic example of how chemical equilibrium principles are applied industrially.
The Haber-Bosch process is used for ammonia production and is a cornerstone of industrial chemistry. This process is modeled on the equilibrium reaction between hydrogen and nitrogen gases to produce ammonia. Achieving a high yield involves optimizing conditions, such as:

• Pressure: Higher pressures favor the formation of ammonia, as the reaction results in fewer moles of gas, shifting equilibrium towards the products.
• Temperature: Although high temperatures increase reaction rates, they can favor the reverse reaction. A compromise is needed for optimal yield.
• Catalysts: Using iron catalysts reduces the energy requirement and accelerates the reaction.

Ammonia is vital for producing fertilizers and various chemicals, demonstrating the critical role of industrial chemistry in meeting global agricultural demands and supporting countless industries.