Question

Carbonyl fluoride, \(\mathrm{COF}_{2}\), is an important intermediate for organic fluorine compounds. It can be prepared by the following reaction: $$\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{CF}_{4}(g) \rightleftharpoons 2 \mathrm{COF}_{2}(g)$$ At \(1000^{\circ} \mathrm{C}, K\) for this reaction is \(0.50 .\) What are the partial pressures of all the gases at equilibrium when the initial partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{CF}_{4}\) are \(0.713 \mathrm{~atm} ?\)

Short answer

Answer: The equilibrium partial pressures are: CO₂: 0.451 atm, CF₄: 0.451 atm, and COF₂: 0.524 atm.

Step-by-step solution

Understand the given information and write the expression for the equilibrium constant.

We are given that initially the partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{CF}_{4}\) are \(0.713\) atm and there is no \(\mathrm{COF}_{2}\) in the system. At equilibrium, we have to find the partial pressures of all the gases. The equilibrium constant, \(K=0.50\). We can write the expression for the equilibrium constant as: $$K = \frac{[\mathrm{COF_2}]^2}{[\mathrm{CO_2}][\mathrm{CF_4}]}$$

Set up the stoichiometric table for the reaction

Let's create a stoichiometry table for the reaction. Denote the change in partial pressure of \(\mathrm{CO}_{2}\) and \(\mathrm{CF}_{4}\) as \(x\)due to the reaction. Initial partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{CF}_{4}\) is \(0.713\) atm. There is initially no \(\mathrm{COF}_{2}\) in the system. $$\begin{array}{c|c|c|c} &[\mathrm{CO_2}] & [\mathrm{CF_4}] &[\mathrm{COF}_2]\\ \hline \text{Initial}& 0.713 & 0.713 & 0\\ \text{Change} & -x & -x & +2x\\ \text{Final}& 0.713-x & 0.713-x & 2x \end{array}$$

Write the expression for the equilibrium constant in terms of x

Now, replace the equilibrium concentrations in terms of x in the equilibrium constant expression: $$0.50 = \frac{(2x)^2}{(0.713-x)(0.713-x)}$$

Solve the equation for x

To solve this equation for x, first, we can simplify it by multiplying both sides by \((0.713-x)^2\): $$(0.50)(0.713-x)^2 = (2x)^2$$ Now, find the value of x: \(x = 0.262\).

Find the partial pressures of all the gases at equilibrium

Now we can use the value of x to find the partial pressures at equilibrium: $$[\mathrm{CO_2}]_{eq} = 0.713 - x = 0.713 - 0.262 = 0.451 \,\mathrm{atm}$$ $$[\mathrm{CF_4}]_{eq} = 0.713 - x = 0.713 - 0.262 = 0.451 \,\mathrm{atm}$$ $$[\mathrm{COF_2}]_{eq} = 2x = 2(0.262) = 0.524 \,\mathrm{atm}$$ So, the partial pressures at equilibrium are: \(\mathrm{CO_2}: 0.451\,\mathrm{atm}\), \(\mathrm{CF_4}: 0.451\,\mathrm{atm}\), and \(\mathrm{COF_2}: 0.524\,\mathrm{atm}\).

Key concepts

Equilibrium Constant

Understanding the equilibrium constant is crucial for grasping chemical equilibrium concepts. It denotes the balance between products and reactants in a reversible reaction once the reaction has reached a state of equilibrium. This constant is usually denoted by the symbol 'K'.

In the context of the given exercise, the equilibrium constant, represented as K = 0.50, indicates the relationship between the partial pressures of the gaseous reactants, CO₂ and CF₄, and the product COF₂, at equilibrium and at a specific temperature of 1000°C. The formula used is:
\[\begin{equation}K = \frac{{[\mathrm{COF_2}]^2}}{{[\mathrm{CO_2}][\mathrm{CF_4}]}}\end{equation}\]
High values of K suggest that the products are favored, while low values indicate a reaction where reactants are favored. In this particular problem, K is 0.50, which suggests a moderate product favorability under given conditions. By manipulating this expression, we can solve for changes in the system and find out the pressures at equilibrium.

Partial Pressure

The concept of partial pressure is essential when dealing with gases involved in chemical reactions, especially in the field of chemical thermodynamics. Partial pressure refers to the pressure that each gas in a mixture would exert if it were alone in the container.

In our exercise, each gas (CO₂, CF₄, and COF₂) contributes to the total pressure of the system. Initially, CO₂ and CF₄ gases have partial pressures of 0.713 atm each. To find out the partial pressures at equilibrium, we utilize the stoichiometric changes as the reaction proceeds and employ the equilibrium constant to establish a relationship between initial and equilibrium states. The partial pressure calculations are intrinsic to predicting how a gas will behave under certain conditions and also in determining the direction a reaction will shift when it is subjected to changes in pressure.

Stoichiometry

Stoichiometry, a section of chemistry that involves the calculation of reactants and products in chemical reactions, plays a foundational role in discovering the quantity of substances needed or produced.

The example problem utilizes stoichiometry in establishing the initial and final quantities of reactants and products. By setting up a table, we account for the stoichiometric coefficients which tell us that for each 'x' amount of reactant that reacts, twice that amount of product is formed due to the reaction's stoichiometry. In the case of the carbonyl fluoride synthesis:
\[\begin{equation}\begin{array}{c|c|c|c} & [\mathrm{CO_2}] & [\mathrm{CF_4}] & [\mathrm{COF}_2]\hline\text{Initial}& 0.713 & 0.713 & 0\text{Change} & -x & -x & +2x\text{Final}& 0.713-x & 0.713-x & 2x\end{array}\end{equation}\]
The 'change' row reflects the stoichiometry of the reaction where the reactants decrease by 'x' and the product increases by '2x'. Overall, stoichiometry is fundamental in predicting the outcomes of chemical reactions and is a necessary tool for scientists working with reactions at equilibrium.