Problem 23 - Chemistry The Science in Context

Question

At equilibrium, the concentrations of gaseous \(\mathrm{N}_{2}, \mathrm{O}_{2},\) and NO in a sealed reaction vessel are \(\left[\mathrm{N}_{2}\right]=3.3 \times 10^{-3} \mathrm{M}\) $$ \left[\mathrm{O}_{2}\right]=5.8 \times 10^{-3} \mathrm{M}, \text { and }[\mathrm{NO}]=3.1 \times 10^{-3} \mathrm{M} . \text { What is } $$ the value of \(K_{\mathrm{c}}\) for the reaction $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$

Step-by-Step Solution

Answer

Answer: The equilibrium constant, Kc, for the reaction is 0.502.

Step 1: Write the Kc expression

The Kc expression for the given reaction is given by: Kc = [NO]^2 / ([N₂] * [O₂])

Step 2: Substitute the given concentrations at equilibrium into the Kc expression

Now, we need to plug in the given equilibrium concentrations into the Kc expression: Kc = ([3.1 * 10^{-3}]^2) / (3.3 * 10^{-3} * 5.8 * 10^{-3})

Step 3: Calculate the value of Kc

In order to calculate the value of Kc, complete the calculation in the expression: Kc = (9.61 * 10^{-6}) / (19.14 * 10^{-6})

Step 4: Simplify and find the value of Kc

Finally, we need to simplify the expression to obtain the value of Kc: Kc = 0.502 The value of Kc for the reaction is 0.502.

Key Concepts

Equilibrium Concentrations

Imagine a crowded room where people are moving around but, as time passes, the amount of people entering and leaving the room becomes equal. This situation is akin to a chemical reaction reaching equilibrium. In chemistry, when a reaction reaches a state where the rate of the forward reaction equals the rate of the reverse reaction, we say it has achieved equilibrium. At this point, the concentrations of reactants and products remain constant over time, not because the reactions have stopped, but because they are occurring at the same rate.

For example, in the textbook exercise, the reaction between nitrogen gas (2) and oxygen gas (2) to form nitrogen monoxide (NO) has reached equilibrium. The given equilibrium concentrations are:

• 2 = 3.3 x 10^-3 M
• 2 = 5.8 x 10^-3 M
• NO = 3.1 x 10^-3 M

These values represent the point at which the formation of NO is balanced by its decomposition back into 2 and 2, leading to no net change in their concentrations.

Reaction Quotient

The measure known as the reaction quotient, denoted by Q, serves as a “snapshot” of a reaction’s status at a given moment. It possesses the same form as the equilibrium constant expression but isn't limited to the state of equilibrium. Instead, it can be calculated using the concentrations of reactants and products at any phase during the reaction.

Q can be compared to the known equilibrium constant (), to predict the reaction's direction:

• If Q < , the reaction will proceed forward, converting reactants into products.
• If Q > , the reaction will go in reverse, forming more reactants from the products.
• If Q = , the reaction is at equilibrium.

This helps chemists determine whether they need to alter conditions to shift the reaction in a desired direction, a concept tied closely with Le Chatelier's principle.

Le Chatelier's Principle

Chemical reactions are a delicate balancing act. Imagine a seesaw that self-adjusts to remain level despite any weight changes. This is the essence of Le Chatelier's principle. It states that if a dynamic equilibrium is disturbed by changing the conditions, such as concentration, temperature, or pressure, the equilibrium will shift to counteract the imposed change, thereby re-establishing equilibrium.

For instance, if more 2 is added to the reaction vessel from our exercise, according to Le Chatelier's principle, the equilibrium would shift to the right (toward the product side) to use up the extra 2, thus producing more NO. Conversely, if the temperature were increased and the reaction was exothermic, the equilibrium would shift to the left to absorb the added heat by producing more reactants. This principle is fundamental for understanding how to control industrial chemical reactions.

Kc Expression

To quantify the equilibrium state of a chemical reaction, we use an equilibrium constant, known as , which is distinct for every reaction at a given temperature. The expression for (the 'c' indicates concentrations are being used) is reflective of the balanced chemical equation and involves concentrations of products over reactants, each raised to the power of their stoichiometric coefficients.

In our exercise, the expression for the reaction 2(g) + 2(g) ⇌ 2 NO(g) would be:

• = [NO]^2 / ([2] * [2])

The square on [NO] is because it has a coefficient of 2 in the balanced equation. To find the equilibrium constant, we substitute the concentrations at equilibrium into this expression. Thus, provides a ratio that helps predict the direction of the reaction and how it will respond to any changes in conditions.