Problem 48 - Chemistry: The Central Science in SI Units, Global Edition

Question

For thereaction, at \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g), K_{c}=55.3\) at \(700 \mathrm{~K}\). In a 10.0-L flask containing an equilibrium mixture of the three gases, there are \(1.30 \mathrm{~g} \mathrm{H}_{2}\) and \(21.0 \mathrm{~g} \mathrm{I}_{2}\). What is the mass of HI in the flask?

Step-by-Step Solution

Answer

The mass of HI in the flask at equilibrium is 10.6 g.

Step 1: Convert mass to moles

We will convert the mass of H2 and I2 to moles using their molar mass. Molar mass of H2 = 2.02g/mol Molar mass of I2 = 253.81g/mol Moles of H2 = mass / molar_mass = 1.30g / 2.02 g/mol = 0.6436 mol Moles of I2 = mass / molar_mass = 21.0g / 253.81 g/mol = 0.0827 mol Moles of HI (initially) = 0

Step 2: Use ICE table

Now we will use the ICE table to find equilibrium moles. H2 + I2 ⇌ 2HI Initial(mol) 0.6436 0.0827 0 Change(mol) -x -x +2x Equil(mol) 0.6436-x 0.0827-x 2x Kc = 55.3 = \([HI]^2 / ([H2][I2])\] We will plug the equilibrium concentrations in the Kc expression and solve for x. 55.3 = \((((2x)^2)/(0.6436-x)(0.0827-x))\)

Step 3: Solve for x

After solving the above equation for x, we get: x = 0.0413 mol Now we can find the moles of HI at equilibrium: Moles of HI at equilibrium = 2x = 2(0.0413) = 0.0826 mol

Step 4: Convert moles of HI to mass

Finally, we will convert the moles of HI back to mass using its molar mass. Molar mass of HI = 127.91g/mol (H=1.01g/mol and I=126.9g/mol) Mass of HI = moles × molar_mass = 0.0826 mol × 127.91 g/mol = 10.6 g The mass of HI in the flask at equilibrium is 10.6 g.

Key Concepts

Understanding Molar Mass

To tackle this exercise, it's important to grasp what molar mass is. Molar mass is the mass of a given substance divided by the amount of substance, measured in g/mol. It allows us to convert between the mass of a substance and the number of moles, which is vital in chemical calculations. For instance:

• Molar mass of diatomic hydrogen (H₂) is 2.02 g/mol because each hydrogen atom has a molar mass of approximately 1.01 g/mol.
• Molar mass of diatomic iodine (I₂) is 253.81 g/mol, as each iodine atom has a molar mass of around 126.9 g/mol.

By knowing molar mass, you can convert mass to moles, a step necessary to reach equilibrium calculations in chemical reactions.

How to Use an ICE Table

The ICE table is a useful tool in chemistry for understanding the equilibrium state of a reaction. ICE stands for Initial, Change, and Equilibrium – the three stages of concentration in a chemical reaction.

• Initial: Start with the initial concentrations (or moles) of the reactants and products. In our problem, initially, hydrogen and iodine have moles, but HI starts at 0.
• Change: During the reaction, concentrations change. We denote this change by 'x', representing how much reactants are converted to products.
• Equilibrium: At equilibrium, we will have expressions for concentrations involving 'x'. Here, for H₂ and I₂ it is (initial moles - x), while for HI it is (2x).

By filling in the ICE table, we create expressions that help us solve for unknown variables like 'x'.

Calculating Equilibrium Concentrations

At equilibrium, the concentrations of reactants and products do not change. These concentrations can be calculated using the ICE table.

• From the example, the equilibrium concentration of hydrogen ( [H₂ ]) and iodine ( [I₂ ]) is (initial moles - x).
• The equilibrium concentration of HI, being the product formed, is 2x, as it forms from each molecule of H₂ and I₂ reacting to make two molecules of HI.

Plugging these values into the equation for the equilibrium constant ( K_c ) allows you to solve for 'x', then determine the amount of each substance at equilibrium.

Understanding the Equilibrium Constant (Kc)

The equilibrium constant, denoted as K_c, is a numeric value that expresses the ratio of the concentrations of the products to the reactants at equilibrium, raised to the power of their stoichiometric coefficients.

• In our exercise, K_c demonstrates how much product (HI) is present at a specified temperature compared to reactants (H₂ and I₂).
• Given the expression K_c = rac{[HI]^2}{[H_2][I_2]} , you can substitute the equilibrium concentrations from the ICE table and solve for unknowns, like 'x'.
• The value of K_c (55.3 in this example) indicates the reaction's position at equilibrium. A large K_c means the products are favored, which, in this case, is HI.

Understanding the equilibrium constant is crucial in predicting and calculating the concentrations of species at equilibrium in a chemical reaction.