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 $$ \mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{COCl}_{2}(\mathrm{g}) $$ has come to equilibrium at \(600 \mathrm{R},\left[\mathrm{Cl}_{2}\right]=\) \(0.00301 \mathrm{mol} / \mathrm{L}\) (a) Calculate the concentrations of \(\mathrm{CO}\) and \(\mathrm{COCl}_{2}\) at equilibrium. (b) Calculate \(K_{\mathrm{r}}\)

Short answer

At equilibrium, \([\text{CO}] = 0.00712\) mol/L and \([\text{COCl}_2] = 0.00308\) mol/L. Equilibrium constant \(K_r\) can be calculated using these concentrations.

Step-by-step solution

Determine Initial and Change in Concentration

First, let's define the initial concentrations and changes as the reaction proceeds to equilibrium for the given reaction: - Initial concentration of \( \text{CO} \) (\( [\text{CO}]_0 = 0.0102 \) mol/L)- Initial concentration of \( \text{Cl}_2 \) (\( [\text{Cl}_2]_0 = 0.00609 \) mol/L)When the reaction reaches equilibrium, the concentration of \( \text{Cl}_2 \) is \( 0.00301 \) mol/L. The change in the concentration of \( \text{Cl}_2 \) is:\[ \Delta[\text{Cl}_2] = [\text{Cl}_2]_0 - [\text{Cl}_2]_{eq} = 0.00609 - 0.00301 = 0.00308 \, \text{mol/L} \]

Calculate the Change in Concentration for CO and COCl2

Given the stoichiometry of the reaction, the change in the concentration of \( \text{CO} \) will be the same as that of \( \text{Cl}_2 \), which is 0.00308 mol/L, and the concentration of \( \text{COCl}_2 \) formed will also be 0.00308 mol/L:- \( \Delta[\text{CO}] = -0.00308 \, \text{mol/L} \)- \( \Delta[\text{COCl}_2] = +0.00308 \, \text{mol/L} \)

Key concepts

Reaction Stoichiometry

In chemistry, stoichiometry is very important to understand how reactants turn into products in a balanced chemical equation. For our problem, the reaction involves carbon monoxide (CO) reacting with chlorine gas (Cl\(_2\)) to form a new compound called phosgene (COCl\(_2\)). This balanced reaction equation is:\[ \text{CO(g)} + \text{Cl}_2\text{(g)} \rightleftharpoons \text{COCl}_2\text{(g)} \]Stoichiometry helps us see that one CO molecule reacts with one Cl\(_2\) molecule to form one COCl\(_2\) molecule.
This one-to-one-to-one ratio is critical because it tells us exactly how the concentration change of one substance will affect the others over time.
To get to equilibrium, both CO and Cl\(_2\) decrease by 0.00308 mol/L, which is the same amount that COCl\(_2\) gains. This clear ratio guides us in predicting how the initial amounts convert to products. With stoichiometry, solving reaction concentration problems becomes a straightforward process.

Equilibrium Concentration

Once a reaction reaches equilibrium, the concentrations of all involved substances remain constant. At equilibrium, the rate of the forward reaction equals the rate of the backward reaction. For our specific reaction, we need to establish the equilibrium concentrations based on the changes from initial conditions:- Initial concentration of CO: 0.0102 mol/L.- Initial concentration of Cl\(_2\): 0.00609 mol/L.- Equilibrium concentration of Cl\(_2\): 0.00301 mol/L.Knowing that Cl\(_2\) dropped by 0.00308 mol/L to reach equilibrium, the change for CO is the same. Thus, the equilibrium concentration for CO is:\[ [\text{CO}]_{eq} = 0.0102 - 0.00308 = 0.00712 \, \text{mol/L} \]For COCl\(_2\), initially, there was none.
It gained what Cl\(_2\) lost, thus its concentration at equilibrium is 0.00308 mol/L.
Once we've established equilibrium concentrations, you can further calculate other properties such as the equilibrium constant.

Equilibrium Constant

Every chemical equilibrium can be described mathematically using the equilibrium constant, symbolized as \( K_r \). This constant gives you the ratio of product concentrations to reactant concentrations when the reaction reaches equilibrium.
For our example:\[ K_{c} = \frac{[\text{COCl}_2]}{[\text{CO}][\text{Cl}_2]} \]Using the equilibrium concentrations we calculated:- \([\text{CO}] = 0.00712\, \text{mol/L}\)- \([\text{Cl}_2] = 0.00301\, \text{mol/L}\)- \([\text{COCl}_2] = 0.00308\, \text{mol/L}\)Plug these values into the equation for \( K_c \):\[K_{c} = \frac{0.00308}{0.00712 \times 0.00301} \approx 0.14\]This value of \( K_c \) tells us the position of equilibrium.
A higher \( K_c \) means more products compared to reactants at equilibrium, while a lower \( K_c \) means more reactants. Understanding the equilibrium constant helps predict the behavior and extent of the reaction under different conditions.