Problem 35 - Chemistry: The Central Science

Question

A mixture of \(0.10 \mathrm{~mol}\) of \(\mathrm{NO}, 0.050 \mathrm{~mol}\) of \(\mathrm{H}_{2}\), and \(0.10 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}\) is placed in a \(1.0-\mathrm{L}\) vessel at \(300 \mathrm{~K}\). The following equilibrium is established: $$ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ At equilibrium [NO \(]=0.062 \mathrm{M}\). (a) Calculate the equilibrium concentrations of \(\mathrm{H}_{2}, \mathrm{~N}_{2}\), and \(\mathrm{H}_{2} \mathrm{O}\). (b) Calculate \(K_{c}\).

Step-by-Step Solution

Answer

The equilibrium concentrations of H2, N2, and H2O are \(0.031 \mathrm{~M}\), \(0.019 \mathrm{~M}\), and \(0.138 \mathrm{~M}\), respectively. The equilibrium constant Kc is 0.157.

Step 1: (a) Calculate Initial Concentrations

First, we will calculate the initial concentrations of NO, H2, and H2O. - For NO: \(c = \frac{n}{V} = \frac{0.10 \mathrm{~mol}}{1.0 \mathrm{~L}} = 0.10 \mathrm{~M}\) - For H2: \(c = \frac{n}{V} = \frac{0.050 \mathrm{~mol}}{1.0 \mathrm{~L}} = 0.050 \mathrm{~M}\) - For H2O: \(c = \frac{n}{V} = \frac{0.10 \mathrm{~mol}}{1.0 \mathrm{~L}} = 0.10 \mathrm{~M}\)

Step 2: (a) Calculate the Change in Concentrations

Let's represent the change in concentration of each species throughout the reaction with x: - For NO: -2x (2 moles of NO are consumed) - For H2: -x (1 mole of H2 is consumed) - For N2: +x (1 mole of N2 is produced) - For H2O: +2x (2 moles of H2O are produced)

Step 3: (a) Calculate the Equilibrium Concentrations

Now, we can express the equilibrium concentrations of NO, H2, N2, and H2O using their initial concentrations and the change in concentrations: - [NO] = 0.10 - 2x M - [H2] = 0.050 - x M - [N2] = x M - [H2O] = 0.10 + 2x M We are given that [NO] = 0.062 M at equilibrium. So, 0.062 = 0.10 - 2x Solve for x: x = 0.019

Step 4: (a) Find the Equilibrium Concentrations for other Species

Now that we have the value of x, we can calculate the equilibrium concentrations for the other species: - [H2] = 0.050 - x = 0.050 - 0.019 = 0.031 M - [N2] = x = 0.019 M - [H2O] = 0.10 + 2x = 0.10 + 2(0.019) = 0.138 M

Step 5: (b) Calculate Kc

We can now calculate the equilibrium constant, Kc, using the equilibrium concentrations of each species: \(K_c = \frac{[\mathrm{N}_{2}][\mathrm{H}_{2}\mathrm{O}]^2}{[\mathrm{NO}]^2[\mathrm{H}_{2}]}\) \(K_c = \frac{(0.019 \mathrm{~M})(0.138 \mathrm{~M})^2}{(0.062 \mathrm{~M})^2(0.031 \mathrm{~M})}\) Kc = 0.157 The equilibrium concentrations of H2, N2, and H2O are 0.031 M, 0.019 M, and 0.138 M, respectively. The equilibrium constant Kc is 0.157.

Key Concepts

Equilibrium Constant

Understanding how a chemical reaction reaches equilibrium is essential, and the equilibrium constant, represented as K_c, is a powerful tool in this regard. Let's think of it as a numerical snapshot of the concentration ratios of products to reactants when a reaction is at a standstill, no longer moving forward or backward. To calculate K_c, one uses the molar concentrations (denoted by square brackets) of the products raised to the power of their coefficients in the balanced chemical equation, divided by those of the reactants in the same manner.

The importance of knowing K_c cannot be understated. It not only gives insights into the relative amounts of products and reactants but also indicates the direction in which a reaction is likely to proceed. A high value suggests products are favored, and conversely, a low value favors reactants. This information is pivotal for predicting the outcome of chemical processes in various fields such as pharmaceuticals, environmental science, and industrial chemistry.

Reaction Quotient

Closely related to the equilibrium constant is the reaction quotient, Q. Think of Q as a 'test' to determine how far a reaction has proceeded at any point in time before reaching equilibrium. To find Q, we apply the same formula used for K_c but using the current concentrations instead of those at equilibrium.

Comparing Q to K_c can forecast the reaction's next steps. If Q < K_c, the reaction will proceed forwards to form more products. If Q > K_c, the reaction will go backwards, forming more reactants. When Q equals K_c, the reaction is at equilibrium, and no further change in concentrations will occur. This comparison is crucial for chemists when manipulating reaction conditions to drive a reaction toward desirable products.

Le Chatelier's Principle

When a system at equilibrium is disturbed, it will adjust to minimize the disturbance—this is the crux of Le Chatelier's Principle. For students grappling with reaction changes, this principle is a reliable guide. It predicts how a reaction responds to alterations such as pressure, temperature, and concentration.

If more reactants are added, the system shifts to produce more products. Conversely, removing a product causes the system to produce more of it. Changing the temperature can be tricky—increasing it for endothermic reactions shifts the equilibrium to the right, producing more product, but it has the opposite effect for exothermic reactions. Pressure changes only affect gaseous reactants, with an increase in pressure shifting the equilibrium towards the side with fewer gas molecules. This principle aids understanding beyond textbook exercises—it's employed in industrial chemical processes to optimize yields.

Molar Concentration

Molar concentration, commonly denoted by a capital 'M' indicating molarity, measures the amount of a substance in a given volume of solution. It tells us how 'concentrated' a solution is and is calculated by dividing the number of moles of solute by the volume of the solution in liters. In our exercise, molar concentration helped in determining the initial amounts of NO, H₂, and H₂O before the reaction began.

In practice, chemists use molar concentration to prepare solutions for experiments, to analyze reaction mixtures, and to predict how reactions will occur in different concentrations. Having precise control over molar concentration allows a chemist to execute reactions efficiently and reproducibly—an invaluable aspect of chemical studies and industry alike.