Questions

To calculate the reaction quotient ($Q$) for the given reaction, we need to consider the concentrations of the gaseous reactants and products. Solid substances are not included in the expression for $Q$.

The balanced chemical equation is: $C(s) + H_2O(g) \rightleftharpoons CO(g) + H_2(g)$

The expression for the reaction quotient $Q$ (or $Q_c$ in terms of concentration) is: $Q = \frac{[CO][H_2]}{[H_2O]}$

Given Information:

• Volume of reaction vessel ($V$) = 2.25 L
• Moles of C ($n_C$) = 6.25 mol (Solid, so not included in $Q$)
• Moles of $H_2O$ ($n_{H_2O}$) = 13.8 mol
• Moles of CO ($n_{CO}$) = 3.70 mol
• Moles of $H_2$ ($n_{H_2}$) = 9.00 mol

Step 1: Calculate the molar concentration (Molarity) of each gaseous species. Concentration ($[X]$) = Moles of X / Volume of vessel

• $[H_2O] = \frac{n_{H_2O}}{V} = \frac{13.8 \text{ mol}}{2.25 \text{ L}} = 6.1333 \text{ M}$
• $[CO] = \frac{n_{CO}}{V} = \frac{3.70 \text{ mol}}{2.25 \text{ L}} = 1.6444 \text{ M}$
• $[H_2] = \frac{n_{H_2}}{V} = \frac{9.00 \text{ mol}}{2.25 \text{ L}} = 4.00 \text{ M}$

Step 2: Substitute the concentrations into the Q expression. $Q = \frac{[CO][H_2]}{[H_2O]}$ $Q = \frac{(1.6444)(4.00)}{(6.1333)}$ $Q = \frac{6.5776}{6.1333}$ $Q \approx 1.07246$

Rounding to two decimal places, based on the least number of significant figures in the concentration values (3.70 mol has 3 sig figs): $Q \approx 1.07$

The reaction quotient $Q$ is 1.07.