Question

For a problem involving the catalyzed reaction of methane and steam, the following reaction table was prepared: $$ \begin{array}{lcccc} \begin{array}{l} \text { Pressure } \\ \text { (atm) } \end{array} & \mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(g) & \Longrightarrow & \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2}(g) \\ \hline \text { Initial } & 0.30 & 0.40 & 0 & 0 \\ \text { Change } & -x & -2 x & +x & +4 x \\ \hline \text { Equilibrium } & 0.30-x & 0.40-2 x & x & 4 x \end{array} $$ Explain the entries in the "Change" and "Equilibrium" rows.

Short answer

Initial pressures decrease by proportions based on stoichiometry. Final equilibrium pressures are the initial minus the change for reactants and initial plus the change for products.

Step-by-step solution

Identify Initial Pressures

Observe that the initial pressures of \(\text{CH}_4\) and \(\text{H}_2O\) are given, whereas the initial pressures of \(\text{CO}_2\) and \(\text{H}_2\) are zero. These values set the starting conditions for the reaction.

Define the Change in Pressure for Reactants

When the reaction proceeds, the pressure of \(\text{CH}_4\) and \(\text{H}_2O\) will decrease. Let \(x\) be the amount by which the pressure of \(\text{CH}_4\) decreases. Since the stoichiometric coefficient for \(\text{H}_2O\) is 2, the change in pressure for \(\text{H}_2O\) will be \(-2x\).

Define the Change in Pressure for Products

For the products, the pressure of \(\text{CO}_2\) will increase by \(x\), and the pressure of \(\text{H}_2\) will increase by \(4x\) due to their stoichiometric coefficients.

Express Equilibrium Pressures

Combine the initial pressures with the changes to express the equilibrium pressures. For \(\text{CH}_4\), the equilibrium pressure is \(0.30 - x\). For \(\text{H}_2O\), it’s \(0.40 - 2x\). For \(\text{CO}_2\) and \(\text{H}_2\), the equilibrium pressures are \((x)\) and \((4x)\) respectively.

Key concepts

Chemical Equilibrium

Chemical equilibrium is the state in which both reactants and products of a chemical reaction are present in concentrations that have no further tendency to change with time. This happens when the forward reaction rate equals the reverse reaction rate. In our example, the equilibrium positions of \(\text{CH}_4\), \(\text{H}_2O\), \(\text{CO}_2\), and \(\text{H}_2\) adjust to a point where their pressures remain constant over time. This means that the amounts of reactants depleting are exactly balanced by the amounts of products being formed.

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. This relationship is derived from the coefficients of the balanced chemical equation. In our example, the reaction of methane (\text{CH}_4) with steam (\text{H}_2O) follows the stoichiometric equation: \(\text{CH}_4(g) + 2 \text{H}_2O(g) → \text{CO}_2(g) + 4 \text{H}_2(g}\). This tells us that one mole of methane reacts with two moles of steam to produce one mole of carbon dioxide and four moles of hydrogen.

Gas Pressure

Gas pressure is the force exerted by gas molecules when they collide with the walls of their container. In chemical reactions involving gases, like our example, changes in gas pressure are observed as reactants are converted to products. The reaction table provided helps us understand these changes. Initially, we have 0.30 atm of \(\text{CH}_4\) and 0.40 atm of \(\text{H}_2O\). As the reaction proceeds, their pressures decrease by \-x\ and \-2x\ respectively. For the products, the pressures increase by \(+x\) for \(\text{CO}_2\) and \(+4x\) for \(\text{H}_2\). These changes are contingent on the stoichiometry of the reaction and help determine equilibrium pressures.