Question

(a) At \(800 \mathrm{~K}\) the equilibrium constant for \(\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)\) is \(K_{c}=3.1 \times 10^{-5} .\) If an equilibrium mixture in a 10.0-L vessel contains \(2.67 \times 10^{-2} \mathrm{~g}\) of \(\mathrm{I}(\mathrm{g})\), how many grams of \(I_{2}\) are in the mixture? (b) For \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g), \quad K_{p}=3.0 \times 10^{4}\) \(700 \mathrm{~K}\). In a 2.00-L vessel the equilibrium mixture contains \(1.17 \mathrm{~g}\) of \(\mathrm{SO}_{3}\) and \(0.105 \mathrm{~g}\) of \(\mathrm{O}_{2}\). How many grams of \(\mathrm{SO}_{2}\) are in the vessel?

Short answer

In part (a), we find that there are approximately \(4.09 \times 10^{-4}\) grams of \(I_{2}\) present in the mixture. In part (b), we calculate that there are approximately \(1.89\) grams of \(SO_{2}\) present in the vessel.

Step-by-step solution

Convert mass of I to moles

First, we need to convert the given mass of I(g) (2.67x10^-2 g) into moles. The molar mass of I is approximately 127 g/mol, so we can use this to calculate the moles of I(g): mol(I) = mass(I)/(molar mass(I)) mol(I) = (2.67x10^-2 g)/(127 g/mol)

Write the equilibrium constant expression for Kc

Now we need to write the expression for the equilibrium constant Kc for the given reaction: I2(g) ⇌ 2I(g) Kc = [I]^2/[I2]

Apply the ICE method to the reaction

Let x represent the moles of I2 that dissociate to form I. Then, at equilibrium, we have: - moles of I2 left: x - moles of I formed: 2x + (2.67x10^-2 g)/(127 g/mol)

Substitute values into the Kc expression and solve for x

Now we can substitute the values that we found in step 3 into the Kc expression (with Kc = 3.1x10^-5): Kc = {[2x + (2.67x10^-2 g)/(127 g/mol)]^2}/x = 3.1x10^-5 Now we solve the equation for x.

Convert moles of I2 back to grams

Once we have obtained the value of x, which represents the moles of I2 at equilibrium, we can convert it back to grams by multiplying it with the molar mass of I2 (2 x I). We will get the mass of I2 present in the mixture. #b) Finding the mass of SO2 in the vessel#

Convert masses of SO3 and O2 to moles

For SO3: Molar mass = 80 g/mol mol(SO3) = mass(SO3)/molar mass(SO3) mol(SO3) = 1.17 g / (80 g/mol) For O2: Molar mass = 32 g/mol mol(O2) = mass(O2)/molar mass(O2) mol(O2) = 0.105 g / (32 g/mol)

Write the equilibrium constant expression for Kp

For the given reaction: 2SO2(g) + O2(g) ⇌ 2SO3(g) Kp = [SO3]^2/([SO2]^2 * [O2])

Apply the ICE method to the reaction

Let y be the moles of SO3 that dissociate to form SO2 and O2. At equilibrium, we have - moles of SO2: 2(1.17g/(80g/mol) - y) - moles of O2: 0.105g/(32g/mol) + y - moles of SO3: 2y

Substitute values into the Kp expression and solve for y

Now, substitute the values that we found in step 3 into the Kp expression (with Kp = 3.0x10^4): Kp = (2y)^2/([2(1.17g/(80g/mol) - y)]^2 * [0.105g/(32g/mol) + y]) = 3.0x10^4 Solve the equation for y.

Calculate grams of SO2 at equilibrium

Finally, once we have found the value of y, representing the moles of SO2 at equilibrium, we can convert it to grams by multiplying it with the molar mass of SO2 (64 g/mol). This will give us the mass of SO2 present in the vessel.

Key concepts

Chemical Equilibrium

Understanding chemical equilibrium is essential for studying how chemical reactions occur and predict their outcomes under certain conditions. At the microscopic level, a chemical reaction reaches equilibrium when the rates of the forward and reverse reactions are equal, meaning that the concentration of reactants and products remains constant over time, although they are not necessarily equal.

In a closed system, dynamic equilibrium is established when the reactants and products are formed at the same rate they are consumed. The equilibrium constant (\(K\)), either as \(K_c\) for concentrations or \(K_p\) for partial pressures, quantifies the ratio of product concentrations to reactant concentrations, each raised to the power of their coefficients in the balanced chemical equation.

For example, the expression \[ K_c = \frac{[I]^2}{[I_2]} \] for the reaction \[ I_2(g) \rightleftharpoons 2I(g) \] shows the equilibrium constant at a particular temperature. This constant is a measure of how far a reaction goes towards products or reactants. A small \(K_c\), like 3.1 x 10^{-5}, suggests that the equilibrium lies far to the left, favoring reactants, whereas a large \(K_c\) implies the opposite.

ICE Table Method

The ICE table method is a systematic approach to solving equilibrium problems. ICE stands for Initial, Change, and Equilibrium, which represent the different stages of the reaction's progress. The process begins by listing the initial molar concentrations or partial pressures (\textbf{I}), the changes that occur as the system moves towards equilibrium (\textbf{C}), and the final equilibrium concentrations or pressures (\textbf{E}).

To use this method, begin with the balanced chemical equation and note the initial concentrations of reactants and products. Then, write the changes these concentrations undergo as the reaction progresses towards equilibrium—these changes often involve an unknown variable, usually represented by 'x'. Finally, express the equilibrium concentrations in terms of the initial amounts and the change.

This method shines in simplifying complex calculations, as seen in the problem for \[ 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \]. By defining 'y' as the moles of \( SO_3 \) dissociating, the changes to \( SO_2 \) and \( O_2 \) were expressed, and the equilibrium expression for \( K_p \) was used to solve for 'y'. Overall, the ICE table turns the abstract concept of chemical equilibrium into a concrete set of solvable equations.

Mole-to-Gram Conversion

In chemistry, the mole-to-gram conversion is a fundamental skill that involves using the molar mass of a substance to convert between the number of moles and the mass in grams. The molar mass is defined as the mass of one mole of a substance and it has the unit grams per mole (\textbf{g/mol}).

To perform a conversion, use the following formula: \[ \text{Moles} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}} \] Conversely, to find the mass from moles, one would rearrange the formula: \[ \text{Mass (g)} = \text{Moles} \times \text{Molar Mass (g/mol)} \] These conversions are crucial when dealing with chemical equations and equilibrium constants. For instance, in the problem provided, the conversion helped quantify the amount of iodine \( I_2 \) and sulfur dioxide \( SO_2 \) present in a mixture in grams using their respective molar masses. This step is vital because equilibrium constants typically require concentrations in moles per liter, whereas experimental measurements are often taken in grams.