Problem 73 - Chemistry The Central Science In Si Units

Question

A mixture of \(\mathrm{CH}_{4}\) and \(\mathrm{H}_{2} \mathrm{O}\) is passed over a nickel catalyst at \(1000 \mathrm{~K}\). The emerging gas is collected in a 5.00-L flask and is found to contain \(8.62 \mathrm{~g}\) of \(\mathrm{CO}, 2.60 \mathrm{~g}\) of \(\mathrm{H}_{2}, 43.0 \mathrm{~g}\) of \(\mathrm{CH}_{4}\), and \(48.4 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\). Assuming that equilibrium has been reached, calculate \(K_{c}\) and \(K_{p}\) for the reaction \(\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g)+3 \mathrm{H}_{2}(g)\).

Step-by-Step Solution

Answer

\(K_c \approx 0.0126\) and \(K_p \approx 10.7\).

Step 1: Calculate Moles of Each Gas

Use the molar masses to convert mass to moles for each gas. The molar mass of CO is "28.01 g/mol", for H2 is "2.02 g/mol", for CH4 is "16.04 g/mol", and for H2O is "18.02 g/mol".\[ n_{CO} = \frac{8.62 \, \text{g}}{28.01 \, \text{g/mol}} = 0.308 \, \text{mol} \]\[ n_{H_2} = \frac{2.60 \, \text{g}}{2.02 \, \text{g/mol}} = 1.287 \, \text{mol} \]\[ n_{CH_4} = \frac{43.0 \, \text{g}}{16.04 \, \text{g/mol}} = 2.68 \, \text{mol} \]\[ n_{H_2O} = \frac{48.4 \, \text{g}}{18.02 \, \text{g/mol}} = 2.69 \, \text{mol} \]

Step 2: Determine Equilibrium Concentrations

To find equilibrium concentrations, divide the moles by the volume of the container (5.00 L):\[ [CO] = \frac{0.308 \, \text{mol}}{5.00 \, \text{L}} = 0.0616 \, \text{M} \]\[ [H_2] = \frac{1.287 \, \text{mol}}{5.00 \, \text{L}} = 0.2574 \, \text{M} \]\[ [CH_4] = \frac{2.68 \, \text{mol}}{5.00 \, \text{L}} = 0.536 \, \text{M} \]\[ [H_2O] = \frac{2.69 \, \text{mol}}{5.00 \, \text{L}} = 0.538 \, \text{M} \]

Step 3: Calculate Equilibrium Constant \(K_c\)

Use the equilibrium concentrations to find \(K_c\). The reaction is \(\mathrm{CH}_4(g) + \mathrm{H}_2O(g) \rightleftharpoons \mathrm{CO}(g) + 3\mathrm{H}_2(g)\). The expression for \(K_c\) is:\[ K_c = \frac{[CO][H_2]^3}{[CH_4][H_2O]} = \frac{(0.0616)(0.2574)^3}{(0.536)(0.538)} \approx 0.0126 \]

Step 4: Calculate Equilibrium Constant \(K_p\)

Convert \(K_c\) to \(K_p\) using the formula: \[ K_p = K_c(RT)^{\Delta n} \]Where \(R = 0.0821 \, \text{L}\cdot\text{atm}/\text{K}\cdot\text{mol}\), \(T = 1000 \, \text{K}\), and \(\Delta n = (1+3)-(1+1) = 2\).\[ K_p = 0.0126 \times (0.0821 \times 1000)^2 \approx 10.7 \]

Key Concepts

Chemical Equilibrium

In any reversible chemical reaction, a state is eventually reached where the rates of the forward and reverse reactions are equal. This is known as chemical equilibrium. During equilibrium, the concentrations of reactants and products no longer change with time, even though the reactions continue to occur. It’s important to understand that equilibrium doesn’t mean the reactions have stopped, but rather that there is a constant interchange of molecules between reactants and products.

A key feature of equilibrium is the equilibrium constant, denoted as \( K_c \) when concentration is used. This constant provides a ratio of product concentrations to reactant concentrations, each raised to the power of their respective coefficients in the balanced chemical equation. It shows how "favorable" a reaction is toward forming products or staying as reactants. For example, a larger \( K_c \) value indicates that, at equilibrium, the products are favored, while a smaller \( K_c \) value suggests the reactants are more dominant. This relationship is critical when predicting the proportions of reactants and products at equilibrium. Chemical equilibrium provides crucial insight into how different conditions can affect the composition of the reaction mixture.

Mole Calculation

Mole calculation is a fundamental aspect of chemistry that involves converting mass into moles. The mole is a basic unit in chemistry that allows chemists to work with the amount of a substance in a convenient way. By using the molar mass of each component, which is the mass of one mole of a substance, you can calculate the number of moles present in a given mass.

For instance, if you have a certain mass of carbon monoxide (as in our example), you would use its molar mass (28.01 g/mol) to convert this mass into moles by using the formula: \[ n = \frac{\text{Mass}}{\text{Molar Mass}}\]This is what we did in the problem when determining how many moles of each gas were present in the flask. Mole calculations are crucial as they provide a stepping stone to determine concentrations used in equilibrium calculations.

It is especially important in chemical equilibrium exercises as it allows conversion of experimentally determined masses into moles, which are necessary for calculating concentration and subsequent equilibrium constants.

Chemical Reaction

A chemical reaction involves the conversion of reactants to products, and it is typically described using a balanced chemical equation. The equation shows the substances involved and their relative quantities, ensuring that mass and number of atoms are conserved during the process.

In the context of the given problem, the reaction is:\[\text{CH}_4(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CO}(g) + 3\text{H}_2(g)\]This shows methane gas reacting with steam to form carbon monoxide and hydrogen gas. The balanced equation tells us that one mole of methane and one mole of water produce one mole of carbon monoxide and three moles of hydrogen. Such equations are vital in understanding how much of each substance is involved and how changes to one side of the reaction affect the other.

Knowing the reaction details allows chemists to calculate equilibrium constants and predict how various factors like temperature and pressure can influence the position of equilibrium. For instance, adjusting concentrations or changing temperature could shift the equilibrium to favor either the products or reactants, in accordance with Le Chatelier's Principle. This understanding is crucial in industrial processes where maximizing yield of a desired product is essential.