Problem 22 - Chemistry: The Central Science in SI Units, Global Edition

Question

Calculate \(K_{c}\) at \(900 \mathrm{~K}\) for \(2 \mathrm{CO}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{C}(s)\) if \(K_{p}=0.0572\) at this temperature.

Step-by-Step Solution

Answer

The equilibrium constant, \(K_c\), for the reaction \(2 CO(g) \rightleftharpoons CO_2(g) + C(s)\) at 900 K is 4.227.

Step 1: Calculate \(\Delta n\) for the reaction

To determine the change in moles of gas, subtract the moles of gaseous reactants from the moles of gaseous products: \[\Delta n = (\text{moles of } CO_2) - (\text{moles of } CO)\] In this reaction, there is 1 mole of \(CO_2\) and 2 moles of \(CO\). So, the change in moles is: \[\Delta n = 1 - 2 = -1\]

Step 2: Calculate \(K_c\) using the relationship between \(K_p\) and \(K_c\)

Now that we know the \(\Delta n\) value, we can use the formula to find \(K_c\): \[K_p = K_c (RT)^{\Delta n}\] We are given that \(K_p = 0.0572\), \(R = 0.0821 \, L \, atm/mol \, K\), and \(T = 900 \, K\). Plug these values into the equation: \[0.0572 = K_c \left( 0.0821 \times 900 \right)^{-1}\] Now, calculate the value inside the parentheses and raise it to the power of \(\Delta n\): \(0.0572 = K_c \times (73.89)^{-1}\) Divide both sides of the equation by \((73.89)^{-1}\) to solve for \(K_c\): \[K_c = 0.0572 \times 73.89 = 4.227\] So, the equilibrium constant \(K_c\) at 900 K for the reaction \(2 CO(g) \rightleftharpoons CO_2(g) + C(s)\) is 4.227.

Key Concepts

Chemical Equilibrium

Chemical equilibrium refers to a state in a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction. This means the concentrations of the reactants and products remain constant over time. At this point, there is no net change in the compositions of the system.
Understanding chemical equilibrium is essential as it allows chemists to predict how a system will respond to different changes such as pressure, temperature, and concentration.

• In a closed system, the reaction does not reach equilibrium immediately when mixed; it may favor reactants or products.
• Equilibrium does not mean equal concentrations, but rather that the rate of creation of reactants equals the rate of creation of products.
• The equilibrium constant (K) helps in determining the extent of a reaction.

K is a constant at a given temperature and can give insight into the position of equilibrium; if K is large, the position of equilibrium is towards the products, indicating a higher concentration of products in equilibrium.

Kc and Kp Relationship

The relationship between equilibrium constants in terms of concentration (\(K_c\)) and pressure (\(K_p\)) is essential for reactions involving gases. This relationship helps in converting between these constants depending on the conditions of the system. It is expressed by the formula:\[ K_p = K_c (RT)^{\Delta n} \]where \(R\) is the universal gas constant (0.0821 L atm/mol K), \(T\) is the absolute temperature, and \(\Delta n\) is the change in moles of gas (moles of gaseous products minus moles of gaseous reactants).
When it comes to gaseous reactions like the one in this problem, calculating the change in moles \(\Delta n\) is crucial. It dictates how the pressure-volume work impacts the position of equilibrium.

• If \(\Delta n = 0\), \(K_c = K_p\).
• If \(\Delta n > 0\), \(K_p > K_c\).
• If \(\Delta n < 0\), \(K_p < K_c\).

In our given reaction, we have \( \Delta n = -1 \), leading us to an essential understanding: \(K_p\) is less than \(K_c\). This relationship showcases the effect of concentration and pressure variations on equilibrium constants.

Gaseous Reactions

Gaseous reactions are a particular type of chemical reaction that involves one or more gases as reactants or products. These reactions are sensitive to changes in pressure and volume, making them unique in how they reach equilibrium.The behavior of gases is significantly influenced by the ideal gas law, \(PV = nRT\), where the state variables pressure (P), volume (V), temperature (T), and number of moles (n) are interconnected.
In the example reaction provided:

• The gases involved are carbon monoxide (\(CO\)) and carbon dioxide (\(CO_2\)).
• Since the reaction involves gaseous species, both \(K_c\) and \(K_p\) can be considered depending on the conditions.
• The presence of the solid carbon (C) does not affect the equilibrium calculations directly, as its concentration is constant.

Understanding gaseous reactions is fundamental to fields like chemistry and engineering, where predicting how gases behave under different conditions can be vital. They exhibit unique properties under different temperatures and pressures, affecting the overall equilibrium of the chemical system.