Analyzing Equilibrium Shift And Partial Pressures In NOCl Decomposition

#h1 Consider the Equilibrium of NOCl Decomposition

This article delves into the equilibrium dynamics of the reversible reaction involving the decomposition of nitrosyl chloride ($NOCl$) into nitric oxide ($NO$) and chlorine gas ($Cl_2$). We will explore how the initial partial pressures of the reactants and products, along with the standard Gibbs free energy change ($\Delta G^0$), influence the equilibrium composition of the reaction mixture. Specifically, we'll examine a scenario where a reaction vessel is filled with 8.06 atm of $NOCl$ and 7.44 atm of $Cl_2$, and analyze the system's behavior as it approaches equilibrium. This analysis will involve calculating the equilibrium constant, determining the direction of the shift to reach equilibrium, and ultimately, quantifying the partial pressures of all species at equilibrium. Understanding these principles is crucial for predicting and controlling chemical reactions in various industrial and environmental applications.

The decomposition of nitrosyl chloride ($NOCl$) into nitric oxide ($NO$) and chlorine gas ($Cl_2$) is a fundamental chemical reaction that provides a valuable model for understanding chemical equilibrium. The reaction is represented by the following equation:

$2 NOCl(g) \rightleftharpoons 2 NO(g) + Cl_2(g) \quad \Delta G^0 = 41. kJ$

The positive value of the standard Gibbs free energy change ($\Delta G^0 = 41. kJ$) indicates that the reaction is non-spontaneous under standard conditions. This means that the reverse reaction, the formation of $NOCl$ from $NO$ and $Cl_2$, is thermodynamically favored under standard conditions. However, the reaction can still proceed in the forward direction under non-standard conditions, such as when the initial partial pressures of the reactants and products deviate significantly from their standard states. To understand the behavior of this system, we must consider the relationship between the Gibbs free energy change, the equilibrium constant, and the reaction quotient.

The Gibbs free energy change ($\Delta G$) is a thermodynamic quantity that determines the spontaneity of a reaction under non-standard conditions. It is related to the standard Gibbs free energy change ($\Delta G^0$), the gas constant (R), the temperature (T), and the reaction quotient (Q) by the following equation:

$\Delta G = \Delta G^0 + RT \ln Q$

The reaction quotient (Q) is a measure of the relative amounts of reactants and products present in a reaction at any given time. For the $NOCl$ decomposition reaction, the reaction quotient is expressed as:

$Q = \frac{P_{NO}^2 P_{Cl_2}}{P_{NOCl}^2}$

where $P_{NOCl}$, $P_{NO}$, and $P_{Cl_2}$ are the partial pressures of $NOCl$, $NO$, and $Cl_2$, respectively. The reaction quotient provides insight into the direction a reversible reaction must shift to reach equilibrium. If Q is less than the equilibrium constant (K), the reaction will shift to the right (towards products) to reach equilibrium. Conversely, if Q is greater than K, the reaction will shift to the left (towards reactants) to reach equilibrium. At equilibrium, $\Delta G = 0$, and the reaction quotient is equal to the equilibrium constant (Q = K).

#h2 Calculating the Equilibrium Constant (K)

The equilibrium constant (K) is a crucial parameter that quantifies the extent to which a reaction will proceed to completion at a given temperature. It is defined as the ratio of products to reactants at equilibrium, with each concentration (or partial pressure for gases) raised to the power of its stoichiometric coefficient. For the $NOCl$ decomposition reaction, the equilibrium constant ($K_p$) in terms of partial pressures is given by:

$K_p = \frac{P_{NO(eq)}^2 P_{Cl_2(eq)}}{P_{NOCl(eq)}^2}$

The equilibrium constant is directly related to the standard Gibbs free energy change ($\Delta G^0$) by the following equation:

$\Delta G^0 = -RT \ln K_p$

where R is the ideal gas constant (8.314 J/mol·K) and T is the temperature in Kelvin. This equation allows us to calculate the equilibrium constant from the standard Gibbs free energy change, which is a tabulated thermodynamic property. For the given reaction, $\Delta G^0 = 41. kJ = 41000 J$. Let's assume the temperature is 298 K (25°C), a common reference temperature for thermodynamic calculations. Substituting these values into the equation, we get:

$41000 J = -(8.314 J/mol·K)(298 K) \ln K_p$

Solving for $ln K_p$:

$\ln K_p = \frac{41000 J}{-(8.314 J/mol·K)(298 K)} \approx -16.55$

Now, we can calculate $K_p$ by taking the exponential of both sides:

$K_p = e^{-16.55} \approx 6.89 \times 10^{-8}$

The small value of $K_p$ indicates that the equilibrium strongly favors the reactants, meaning that at equilibrium, the partial pressure of $NOCl$ will be significantly higher than the partial pressures of $NO$ and $Cl_2$. This is consistent with the positive $\Delta G^0$ value, which suggests that the reverse reaction is favored under standard conditions. The magnitude of the equilibrium constant provides valuable information about the relative amounts of reactants and products at equilibrium. A large K value (K >> 1) indicates that the products are favored at equilibrium, while a small K value (K << 1) indicates that the reactants are favored. In this case, the extremely small $K_p$ value suggests that very little $NOCl$ will decompose into $NO$ and $Cl_2$ at equilibrium. Understanding the equilibrium constant is essential for predicting the extent of a reaction and optimizing reaction conditions to maximize product yield.

#h2 Determining the Direction of Shift to Reach Equilibrium

To predict the direction in which the reaction will shift to reach equilibrium, we need to compare the initial reaction quotient (Q) with the equilibrium constant ($K_p$). The reaction quotient (Q) is calculated using the initial partial pressures of the reactants and products. In this scenario, the reaction vessel is initially filled with 8.06 atm of $NOCl$ and 7.44 atm of $Cl_2$. There is no $NO$ initially present, so its partial pressure is 0 atm.

The initial reaction quotient ($Q_p$) is calculated as follows:

$Q_p = \frac{P_{NO(initial)}^2 P_{Cl_2(initial)}}{P_{NOCl(initial)}^2} = \frac{(0)^2 (7.44)}{(8.06)^2} = 0$

Comparing $Q_p$ with $K_p$, we have:

$Q_p = 0 < K_p = 6.89 \times 10^{-8}$

Since $Q_p$ is less than $K_p$, the reaction will shift to the right, favoring the formation of products ($NO$ and $Cl_2$), to reach equilibrium. This is because the initial conditions have a lower ratio of products to reactants compared to the equilibrium conditions. The system will adjust by converting some of the $NOCl$ into $NO$ and $Cl_2$ until the ratio of products to reactants matches the equilibrium constant. The magnitude of the difference between $Q_p$ and $K_p$ provides an indication of the extent to which the reaction will shift. In this case, the significant difference between 0 and $6.89 \times 10^{-8}$ suggests that a considerable amount of $NOCl$ will decompose, even though the equilibrium constant is small. This highlights the importance of considering both the equilibrium constant and the initial conditions when predicting the behavior of a reversible reaction.

#h2 Calculating Partial Pressures at Equilibrium

To calculate the partial pressures of all species at equilibrium, we can use an ICE table (Initial, Change, Equilibrium). Let's define 'x' as the change in partial pressure of $Cl_2$ as the reaction proceeds towards equilibrium.

At equilibrium, the partial pressures are related by the equilibrium constant expression:

$K_p = \frac{P_{NO(eq)}^2 P_{Cl_2(eq)}}{P_{NOCl(eq)}^2} = \frac{(2x)^2 (7.44 + x)}{(8.06 - 2x)^2} = 6.89 \times 10^{-8}$

Since $K_p$ is very small, we can assume that 'x' is much smaller than 8.06 and 7.44, simplifying the equation:

$6.89 \times 10^{-8} \approx \frac{(4x^2)(7.44)}{(8.06)^2}$

Solving for $x^2$:

$x^2 \approx \frac{(6.89 \times 10^{-8})(8.06)^2}{4(7.44)} \approx 1.50 \times 10^{-7}$

Taking the square root:

$x \approx \sqrt{1.50 \times 10^{-7}} \approx 3.87 \times 10^{-4}$ atm

Now we can calculate the equilibrium partial pressures:

$P_{NOCl(eq)} = 8.06 - 2x \approx 8.06 - 2(3.87 \times 10^{-4}) \approx 8.06$ atm

$P_{NO(eq)} = 2x \approx 2(3.87 \times 10^{-4}) \approx 7.74 \times 10^{-4}$ atm

$P_{Cl_2(eq)} = 7.44 + x \approx 7.44 + 3.87 \times 10^{-4} \approx 7.44$ atm

As we can see, the change in partial pressures is very small due to the small value of $K_p$. The equilibrium partial pressures of $NOCl$ and $Cl_2$ are essentially the same as their initial partial pressures, while the equilibrium partial pressure of $NO$ is very low. This confirms that the equilibrium strongly favors the reactants under these conditions. The small change in partial pressures justifies our initial assumption that 'x' is much smaller than 8.06 and 7.44. If this assumption were not valid, we would need to solve the quadratic equation without simplification, which would be a more complex calculation. The ICE table method, coupled with appropriate simplifying assumptions, is a powerful tool for analyzing chemical equilibria and determining the composition of reaction mixtures at equilibrium.

#h2 Conclusion

In conclusion, this analysis of the $NOCl$ decomposition equilibrium demonstrates the interplay between thermodynamics and kinetics in determining the composition of a reaction mixture. The positive standard Gibbs free energy change ($\Delta G^0$) indicates that the reaction is non-spontaneous under standard conditions, resulting in a small equilibrium constant ($K_p$). The comparison of the initial reaction quotient ($Q_p$) with $K_p$ revealed that the reaction would shift to the right to reach equilibrium, but the small value of $K_p$ meant that the extent of this shift was limited. The calculated equilibrium partial pressures confirmed that the equilibrium strongly favors the reactants, with very little $NOCl$ decomposing into $NO$ and $Cl_2$. This example illustrates the importance of understanding equilibrium principles in predicting the behavior of chemical reactions and optimizing reaction conditions for desired outcomes. The concepts discussed here are fundamental to various fields, including chemical engineering, environmental science, and materials science, where controlling chemical reactions is essential.