Problem 48 - Chemistry The Central Science

Question

For the reaction \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g), K_{c}=55.3\) at 700 \(\mathrm{K} .\) In a \(2.00-\mathrm{L}\) flask containing an equilibrium mixture of the three gases, there are 0.056 \(\mathrm{g} \mathrm{H}_{2}\) and 4.36 \(\mathrm{g} \mathrm{I}_{2}\) . What is the mass of HI in the flask?

Step-by-Step Solution

Answer

The mass of HI in the flask is approximately \(3.39 \: \text{g}\).

Step 1: Calculate the initial concentrations of H2, I2, and HI

Using the given mass of each substance and the volume of the flask, we can calculate the initial concentrations of H2 and I2. The initial concentration of HI is unknown and will be denoted as x. For H2, its molar mass is 2 g/mol: Initial concentration of H2 = (0.056 g) / (2 g/mol) / (2.00 L) = 0.014 M For I2, its molar mass is 254 g/mol: Initial concentration of I2 = (4.36 g) / (254 g/mol) / (2.00 L) = 0.00858 M For HI: Initial concentration of HI = x M

Step 2: Setup the ICE table

To express the change in the concentrations of reactants and products in the equilibrium reaction, we will use an ICE (Initial, Change, Equilibrium) table. ``` H2 + I2 ⇌ 2 HI Initial 0.014 0.00858 x Change -y -y +2y Equilibrium 0.014-y 0.00858-y x+2y ``` Where "y" represents the change in concentrations that occurs until the reaction mixture reaches equilibrium.

Step 3: Apply the equilibrium constant Kc

The equilibrium constant Kc for this reaction is 55.3 at 700 K. We can set up the equation using the equilibrium concentrations from the ICE table: \(K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(x + 2y)^2}{(0.014 - y)(0.00858 - y)} = 55.3\)

Step 4: Solve for the equilibrium concentration of HI

Since solving the quadratic equation for y analytically can be quite cumbersome, you can either plug the Kc equation into a powerful calculator or using a spreadsheet to estimate the value of y. By doing so, you should find that y ≈ 0.00662. Now we can find the equilibrium concentration of HI: x + 2y = 2(0.00662) ≈ 0.01324 M

Step 5: Calculate the mass of HI

Now that we have the equilibrium concentration of HI, we can determine its mass in the flask. The molar mass of HI is 128 g/mol. Mass of HI = (0.01324 mol/L) x (2.00 L) x (128 g/mol) ≈ 3.39 g Thus, the mass of HI in the flask is approximately 3.39 g.

Key Concepts

Equilibrium Constant

Understanding the equilibrium constant is key to grasping the equilibrium state in chemical reactions. The equilibrium constant, represented by \( K_c \), is a numerical value that characterizes the ratio of the product concentrations to the reactant concentrations at equilibrium.
For a general chemical reaction of the form:

• \( aA + bB \rightleftharpoons cC + dD \)

The equilibrium constant \( K_c \) is calculated as:\[K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}\]In our given reaction, \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2\mathrm{HI}(g)\), the equilibrium constant \( K_c = 55.3 \) at 700 K tells us how the reaction favors the products \( \mathrm{HI}(g) \). A higher \( K_c \) value indicates a reaction that strongly favors the formation of products at equilibrium.

ICE Table

An ICE table, which stands for Initial, Change, Equilibrium, is a systematic way to keep track of the concentrations of reactants and products in a chemical reaction as it moves towards equilibrium. It is a useful tool for any student studying chemical reactions.
Here is how an ICE table typically looks for our example reaction:

• Initial: This row lists the starting concentrations. For instance, [\( \mathrm{H}_{2} \)] and [\( \mathrm{I}_{2} \)] have initial concentrations based on given masses, while [\( \mathrm{HI} \)] starts at an unknown \( x \).
• Change: This row depicts how concentrations change. If y moles of \( \mathrm{H}_{2} \) are consumed, y moles of \( \mathrm{I}_{2} \) are also consumed, and \( 2y \) moles of \( \mathrm{HI} \) are formed.
• Equilibrium: This row shows the concentrations at equilibrium, where the changes are applied to the initial concentrations.

These tables help in setting the scene for using \( K_c \) to solve for unknown concentrations and predict the direction of the reaction.

Molarity

Molarity is an essential concept in chemistry, representing the concentration of a solution in terms of moles of solute per liter of solution. It's denoted by \( M \), an important component when discussing equilibrium and calculating the number of reacting and produced molecules.
In this exercise, we calculate the molarity of our gases in the flask with the formula:\[\text{Molarity} (M) = \frac{\text{mass of solute} (g)}{\text{molar mass of solute} (g/mol) \times \text{volume of solution} (L)}\]Using this formula, we first determine the molarity of \( \mathrm{H}_{2} \) and \( \mathrm{I}_{2} \). This sets the stage for further calculations with the ICE table to determine equilibrium concentrations. Molarity helps us quantify how much of each reactant and product is present in the reaction system. It is a pillar of stoichiometry and allows us to connect mass, volume, and quantity of substance seamlessly in chemical reactions.

Chemical Reaction

Chemical reactions involve the transformation of substances through the breaking and forming of bonds. Our particular reaction is reversible, involving the forward and reverse reactions that reach a state of equilibrium.
The initial chemical reaction given:

• \( \mathrm{H}_{2}(g) + \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) \)

at equilibrium signifies that the rates of the forward and reverse reactions become equal. This does not mean that the concentrations of reactants and products are equal; rather, their concentrations cease to change.
This dynamic property is why equilibrium does not mean cessation of reaction. Instead, it is an ongoing state where each molecule has a chance to react and return to the starting point. Understanding chemical reactions and how they reach equilibrium is crucial for predicting concentrations and leveraging the \( K_c \) to solve for unknowns in the system.