Question

At high temperatures, elemental nitrogen and oxygen react with each other to form nitrogen monoxide. $${\mathrm{N}}_2(g)+{\mathrm{O}}_2(g)\rightleftharpoons 2\mathrm{NO}(g)$$ Suppose the system is analyzed at a particular temperature, and the equilibrium concentrations are found to be $\left[{\mathrm{N}}_2\right]=0.041\mathrm{M},\left[{\mathrm{O}}_2\right]=0.0078M,$ and $[\mathrm{NO}]=4.7\times {10}^{-4}M.$ Calculate the value of $K$ for the reaction.

Short answer

The value of the equilibrium constant (K) for the given reaction can be calculated using the formula $$K=\frac{[\mathrm{NO}{]}^2}{[{\mathrm{N}}_2]\times [{\mathrm{O}}_2]}$$. Plugging in the given equilibrium concentrations, we have $$K=\frac{(4.7\times {10}^{-4}{)}^2}{(0.041)(0.0078)}\approx 6.91\times {10}^{-4}$$.

Step-by-step solution

Write the Formula for the Equilibrium Constant (K)

To find the equilibrium constant K, we will use the formula: $$K=\frac{[\mathrm{Products}]}{[\mathrm{Reactants}]}$$ In this reaction, the product is NO and the reactants are N₂ and O₂. The balanced chemical equation is: $${\mathrm{N}}_2(g)+{\mathrm{O}}_2(g)\rightleftharpoons 2\mathrm{NO}(g)$$

Plug in the Given Equilibrium Concentrations to the Formula

: Plug the given concentrations of N₂, O₂, and NO into the K equation: $$K=\frac{[\mathrm{NO}{]}^2}{[{\mathrm{N}}_2]\times [{\mathrm{O}}_2]}$$ Given data: $$[{\mathrm{N}}_2]=0.041\mathrm{M}$$ $$[{\mathrm{O}}_2]=0.0078\mathrm{M}$$ $$[\mathrm{NO}]=4.7\times {10}^{-4}\mathrm{M}$$

Calculate K

: Substitute the given concentrations into the formula and solve for K: $$K=\frac{(4.7\times {10}^{-4}{)}^2}{(0.041)(0.0078)}$$ $$K=\frac{2.209\times {10}^{-7}}{3.198\times {10}^{-4}}$$

Simplify the Expression and Find the Value of K

: Simplify and find K: $$K\approx 6.91\times {10}^{-4}$$ So, the value of the equilibrium constant (K) for the given reaction is approximately $6.91\times {10}^{-4}$.

Key concepts

Chemical Equilibrium

When a chemical reaction occurs, it often reaches a state where the rate of the forward reaction equals the rate of the reverse reaction, meaning that the concentrations of the reactants and products remain constant over time. This state is known as chemical equilibrium.

Imagine two children on a seesaw who are in perfect balance; even if they move up and down, they don't tip the seesaw over to one side permanently. Similarly, in a chemical system at equilibrium, the quantities of reactants and products don't change anymore, even though their molecules are still reacting with each other. A key point here is that equilibrium doesn't mean the reactants and products are in equal amounts; it simply means their ratios don't change. Understanding equilibrium is crucial for predicting how chemicals react in a variety of environments, including biological systems, industrial processes, and in the natural world.

Equilibrium Concentrations

The equilibrium concentrations of the reactants and products are the amounts of these substances at the precise moment the reaction reaches equilibrium. These concentrations are denoted by square brackets, for example, $${\mathrm{N}}_2$$ for nitrogen. To extend our seesaw analogy, these concentrations would represent the relative positions of our children on the seesaw when they've stopped moving — they've found a stable position.

Calculating these concentrations can sometimes be complex, especially when not all the initial concentrations of reactants and products are known. In such cases, chemists use the concept of the reaction quotient (which we'll discuss next) to move forward. With the equilibrium concentrations, however, as shown in our exercise, you can directly calculate the equilibrium constant—a numerical value that gives us profound insight into the reaction's behavior.

Reaction Quotient

The reaction quotient (Q) expresses the ratio of the concentrations of the products to the reactants at any point during a reaction, not just at equilibrium.

While the equilibrium constant (K) is like a fixed address telling you where you're going, the reaction quotient is akin to your current GPS location, which tells you where you are in relation to your destination. By comparing Q to K, chemists can predict which direction the reaction must shift to reach equilibrium.

• If $Q<K$), the reaction moves forward (toward the products).
• If $Q>K$), the reaction moves backward (toward the reactants).
• If $Q=K$), the reaction is already at equilibrium.

The concept of Q is valuable when you don't know if the system is at equilibrium or if you have initial concentrations instead of equilibrium concentrations. It plays a pivotal role in understanding and predicting the progression of chemical reactions.