Problem 11 - Chemistry and Chemical Reactivity

Question

A mixture of \(\mathrm{CO}\) and \(\mathrm{Cl}_{2}\) is placed in a reaction flask: \([\mathrm{CO}]=0.0102 \mathrm{mol} / \mathrm{L}\) and \(\left[\mathrm{Cl}_{2}\right]=0.00609 \mathrm{mol} / \mathrm{L}\) When the reaction $$ \mathbf{C O}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \operatorname{COCl}_{2}(\mathrm{g}) $$ has come to equilibrium at \(600 \mathrm{K},\left[\mathrm{Cl}_{2}\right]=\) \(0.00301 \mathrm{mol} / \mathrm{L}\) (a) Calculate the concentrations of \(\mathrm{CO}\) and \(\mathrm{COCl}_{2}\) at equilibrium. (b) Calculate \(K_{c}\).

Step-by-Step Solution

Answer

[CO] = 0.00712 mol/L; [COCl2] = 0.00308 mol/L; Kc = 140.04.

Step 1: Identify Initial and Change in Concentration

First, identify the initial concentrations for CO and Cl_{2} and the change in concentration for Cl_{2}. Initially, [Cl_{2}] = 0.00609 \, \text{mol/L} and [CO] = 0.0102 \, \text{mol/L}. At equilibrium, \[ [Cl_{2}] = 0.00301 \, \text{mol/L}\].The change in Cl_{2} is thus \[ \Delta \mathrm{Cl_{2}} = 0.00301 - 0.00609 = -0.00308 \, \text{mol/L} \].

Step 2: Determine Change for Other Reactants and Products

The stoichiometry of the reaction is 1:1:1. Therefore, the change in concentration for CO is also \(-0.00308 \mathrm{mol/L} \) while for COCl_{2}, it is \(+0.00308 \, \text{mol/L}\). Subsequently, we determine:For CO:\[ [\mathrm{CO}]_{eq} = [\mathrm{CO}]_{init} - \Delta \mathrm{CO} = 0.0102 - 0.00308 = 0.00712 \, \text{mol/L} \].For COCl_{2}:\[ [\mathrm{COCl}_{2}]_{eq} = 0 + 0.00308 = 0.00308 \, \text{mol/L} \].

Step 3: Calculate Equilibrium Constant \(K_c\)

Write the expression for the equilibrium constant \(K_c\) as:\[ K_c = \frac{[\mathrm{COCl}_{2}]_{eq}}{[\mathrm{CO}]_{eq}[\mathrm{Cl}_{2}]_{eq}} \].Substitute the equilibrium concentrations:\[ K_c = \frac{0.00308 \, \mathrm{mol/L}}{0.00712 \, \mathrm{mol/L} \times 0.00301 \, \mathrm{mol/L}} \].Carry out the calculation to find the value of \(K_c\).

Key Concepts

Reaction Stoichiometry

Understanding reaction stoichiometry is crucial for mastering chemical equilibrium problems. It involves examining the balanced chemical equation to discern how reactants transform into products. In our equation, \[\mathrm{CO(g) + Cl_2(g) \rightleftharpoons COCl_2(g)},\] the stoichiometric coefficients are all 1. This indicates that one molecule of CO reacts with one molecule of \(\mathrm{Cl_2}\) to produce one molecule of \(\mathrm{COCl_2}\). When determining concentration changes, these coefficients guide us. For instance, the reduction in \(\mathrm{Cl_2}\) by \(-0.00308 \, \text{mol/L}\) entails the same decrease for CO, because of their 1:1 ratio.

Furthermore, the formation of \(\mathrm{COCl_2}\) also corresponds directly to these coefficients, increasing by the same magnitude, \(0.00308 \, \text{mol/L}\). Reaction stoichiometry serves as a map, tying together the changes of each substance involved over the course of the reaction.

Equilibrium Concentrations

Equilibrium concentrations show us the lasting state of reactants and products after the chemical reaction reaches balance. This state is characterized by no further macroscopic change in concentration over time. Let’s see how to determine them for our reaction.

• The original concentrations given are \(\mathrm{[CO] = 0.0102 \, mol/L}\) and \(\mathrm{[Cl_2] = 0.00609 \, mol/L}\).


• Since \(\mathrm{[Cl_2]}\) drops to \(0.00301 \, \text{mol/L}\) at equilibrium, we calculate CO's equilibrium concentration by subtracting the change from its starting value: \(\mathrm{[CO]_{eq} = 0.0102 \, mol/L - 0.00308 \, mol/L = 0.00712 \, mol/L}\).
• Simultaneously, the concentration of \(\mathrm{COCl_2}\) produced is exactly equal to the decrease in CO, thus \(\mathrm{[COCl_2]_{eq} = 0.00308 \, mol/L}\).

Equilibrium concentrations perfectly illustrate how the system stabilizes. They are essential for enabling further calculations and for understanding reaction dynamics on a deeper level.

Chemical Equilibrium Calculations

Performing chemical equilibrium calculations involves using equilibrium concentrations to quantify the reaction's balance point. The equilibrium constant \(K_c\) offers insight into the position of equilibrium and is crucial for predicting reaction behavior.

To find \(K_c\), write the expression:\[K_c = \frac{\mathrm{[COCl_2]_{eq}}}{\mathrm{[CO]_{eq} [Cl_2]_{eq}}}.\]
Insert our values:\[K_c = \frac{0.00308 \, \text{mol/L}}{0.00712 \, \text{mol/L} \times 0.00301 \, \text{mol/L}}.\]
Calculate this ratio to derive \(K_c\). A larger \(K_c\) implies a significant formation of products at equilibrium, while a smaller \(K_c\) indicates a reaction favoring the reactants.

Understanding \(K_c\) is vital because it tells us about the extent of a reaction and provides a foundation for anticipating how changes in conditions affect the system. With each calculation, students grasp more about the nature of chemical equilibria.