Problem 71 - Chemistry and Chemical Reactivity

Question

Characterize each of the following as product- or reactant-favored at equilibrium. $$\begin{aligned} &\text { (a) } \mathrm{CO}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{g})\\\ &&K_{\mathrm{p}}=1.2 \times 10^{45} \end{aligned}$$ $$\begin{aligned} &\text { (b) } \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g})\\\ &&K_{\mathrm{p}}=9.1 \times 10^{-41} \end{aligned}$$ $$\begin{aligned} &\text { (c) } \mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{COCl}_{2}(\mathrm{g})\\\ &&K_{\mathrm{p}}=6.5 \times 10^{11} \end{aligned}$$

Step-by-Step Solution

Answer

(a) Product-favored, (b) Reactant-favored, (c) Product-favored.

Step 1: Understand the Equilibrium Constant

For each of the given reactions, the equilibrium constant, \( K_p \), provides insight into the favorability of either reactants or products. A very large \( K_p \) (much greater than 1) suggests that the products are favored at equilibrium, while a very small \( K_p \) (much less than 1) indicates that the reactants are favored.

Step 2: Analyze Reaction (a)

In reaction (a), "\( \mathrm{CO}( ext{g}) + 1/2 \mathrm{O}_{2}( ext{g}) \rightleftharpoons \mathrm{CO}_{2}( ext{g}) \)" with \( K_p = 1.2 \times 10^{45} \), since the equilibrium constant is extremely large, this indicates the reaction is product-favored. This means that at equilibrium, nearly all the reactants have been converted to products.

Step 3: Analyze Reaction (b)

In reaction (b), "\( \mathrm{H}_{2}\mathrm{O}( ext{g}) \rightleftharpoons \mathrm{H}_{2}( ext{g}) + 1/2 \mathrm{O}_{2}( ext{g}) \)" with \( K_p = 9.1 \times 10^{-41} \), the very small equilibrium constant indicates that this reaction is reactant-favored. This means at equilibrium, very little of the reactants convert to products.

Step 4: Analyze Reaction (c)

In reaction (c), "\( \mathrm{CO}( ext{g}) + \mathrm{Cl}_2( ext{g}) \rightleftharpoons \mathrm{COCl}_{2}( ext{g}) \)" with \( K_p = 6.5 \times 10^{11} \), the large value of \( K_p \) suggests the reaction is product-favored. Like reaction (a), this means products predominate at equilibrium.

Key Concepts

Equilibrium Constant

The equilibrium constant, often symbolized as\( K \), is a vital parameter in chemical reactions at equilibrium. What it essentially does is measure the ratio of concentrations of products to reactants, each raised to the power of their stoichiometric coefficients from the balanced equation. When \( K \) is determined to be much larger than 1, the concentration of products is significantly higher than that of reactants at equilibrium. Conversely, if \( K \) is much less than 1, the reactants are predominant. This constant doesn't care about initial concentrations or dynamics such as temperature, pressure (except for \( Kp \)), or volume directly; it only describes the extent of reaction. Just remember, the larger the \( K \), the more product-favored the reaction is at equilibrium.

Product-Favored Reaction

When a reaction is described as 'product-favored', this means that the products form in much larger quantities compared to the reactants once the reaction reaches equilibrium. This scenario corresponds with having a very large value of the equilibrium constant, \( K \). Take reaction (a) from our exercise:

• Its \( Kp \) value is given as \( 1.2 \times 10^{45} \), a tremendously large number!
• Because of this high \( Kp \) value, almost all the reactants convert into products by the time equilibrium is achieved.

In essence, a product-favored reaction is one where the outcome sees the balance tipped strongly towards the side of the products.

Reactant-Favored Reaction

On the flip side, a reactant-favored reaction presents a case where the reactants are more plentiful compared to the products at equilibrium. In contrast to our previous section, reactant-favored reactions have very small \( K \) values. An excellent example is reaction (b) of our exercise:

• The \( Kp \) for the reaction is \( 9.1 \times 10^{-41} \), which is minuscule.
• Such a small \( Kp \) indicates that the equilibrium lies far to the left, meaning not many products form.

So, when you encounter a small \( K \) value, you can be fairly confident that the amount of product is negligible compared to reactants.

Kp (Equilibrium Constant for Gases)

The equilibrium constant for reactions involving gases is specifically referred to as \( Kp \). This helps us measure the partial pressures of gaseous reactants and products at equilibrium. Unlike its counterpart, \( Kc \), which uses concentration, \( Kp \) is used for reactions where gases are involved. Here's how it works:

• In reactions like (a) and (c), where gases are the primary substances involved, one must use pressure instead of concentrations.
• For instance, \( Kp \) in reaction (c) is \( 6.5 \times 10^{11} \), a huge number that tells us the products outweigh the reactants.

Keep in mind that the ideal gas law connects the two (\( Kc \) and \( Kp \)), but \( Kp \) stands singularly helpful when dealing with reactions in a gaseous state.