Question

Consider the following equilibrium between oxides of nitrogen $$3 \mathrm{NO}(g) \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g)$$ (a) Use data in Appendix C to calculate \(\Delta H^{\circ}\) for this reaction. (b) Will the equilibrium constant for the reaction increase or decrease with increasing temperature? (c) At constant temperature, would a change in the volume of the container affect the fraction of products in the equilibrium mixture?

Short answer

(a) The standard enthalpy change for this reaction is -55.99 kJ/mol. (b) The equilibrium constant for the reaction will decrease with increasing temperature. (c) At constant temperature, changing the volume of the container would affect the fraction of products in the equilibrium mixture. Decreasing the volume will favor the formation of products, while increasing the volume will favor the formation of reactants.

Step-by-step solution

(a) Calculate the Standard Enthalpy Change for the Reaction

To calculate ΔHº for the given reaction, we need to find the standard enthalpies of formation for all reactants and products. From Appendix C, we have: 1. The standard enthalpy of formation for NO(g): ΔHfº(NO(g)) = 90.25 kJ/mol 2. The standard enthalpy of formation for NO₂(g): ΔHfº(NO₂(g)) = 33.18 kJ/mol 3. The standard enthalpy of formation for N₂O(g): ΔHfº(N₂O(g)) = 81.57 kJ/mol Now, we can use the following equation to calculate the standard enthalpy change for the reaction: \[ ΔHº_\text{reaction} =Σ(\text{moles of products} × ΔHfº_\text{products}) - Σ(\text{moles of reactants} × ΔHfº_\text{reactants}) \] For our reaction, this is: \[ ΔHº_\text{reaction} = [(1 × ΔHfº(NO₂(g))) + (1 × ΔHfº(N₂O(g)))] - [(3 × ΔHfº(NO(g)))] \] Plugging in the values and calculating: \( ΔHº = [(1 × 33.18) + (1 × 81.57)] - [(3 × 90.25)] = -55.99 \, kJ/mol \) (a) The standard enthalpy change for this reaction is -55.99 kJ/mol.

(b) Effect of Temperature on the Equilibrium Constant

To determine the effect of increasing temperature on the equilibrium constant, we can use the van't Hoff equation, which relates the enthalpy change, ΔHº, the equilibrium constant, K, and the temperature, T as follows: \[ \frac{d(\ln{K})}{dT} = \frac{ΔHº}{RT^2} \] Given the standard enthalpy change, ΔHº = -55.99 kJ/mol (exothermic), we can analyze its effect on the equilibrium constant as temperature increases. For an exothermic reaction, ΔHº is negative. According to the equation, if the temperature increases, the value of the derivative d(ln K)/dT will be negative, implying that the equilibrium constant K will decrease. (b) The equilibrium constant for the reaction will decrease with increasing temperature.

(c) Effect of Volume Change on the Equilibrium Mixture

To analyze the effect of volume change on the equilibrium mixture at constant temperature, we can use Le Chatelier's principle. It states that if a system at equilibrium is subjected to a stress (such as a change in volume), the system will adjust to oppose the stress and restore a new equilibrium. For the given reaction: \[ 3 \, \mathrm{NO}(g) \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g) \] We can see that the reaction has 3 moles of reactants (3 NO) and 2 moles of products (1 NO₂ + 1 N₂O). Therefore, decreasing the volume of the container would favor the side with fewer moles, in this case, the product side. Conversely, increasing the volume would favor the side with more moles, the reactant side. Thus, the effect of volume change on the equilibrium mixture at constant temperature depends on the direction of the volume change. (c) At constant temperature, changing the volume of the container would affect the fraction of products in the equilibrium mixture. Decreasing the volume will favor the formation of products, while increasing the volume will favor the formation of reactants.

Key concepts

Le Chatelier's Principle

Le Chatelier's Principle is an essential concept in understanding how changes in a system's conditions can affect its equilibrium. When a dynamic equilibrium is disturbed by changing the conditions such as pressure, temperature, or concentration, the system responds to counteract the change and restore equilibrium.

In the context of volume changes, the principle helps us predict the behavior of gaseous reactions:

• If the volume decreases, the pressure increases, and the system will shift to the side with fewer gas moles.
• If the volume increases, the pressure decreases, and the system shifts to the side with more gas moles.

For the reaction involving NO, NO₂, and N₂O, decreasing volume favors the formation of NO₂ and N₂O as it shifts the equilibrium towards the side with fewer moles.

Standard Enthalpy Change

Standard Enthalpy Change (\( ΔH^{ ext{°}} \)) is a measure of the heat absorbed or released during a chemical reaction at constant pressure and under standard conditions (1 atm and 25°C). It provides insight into the energy dynamics of a reaction.

To calculate \( ΔH^{ ext{°}} \), use the formula:

• \[ΔH^{ ext{°}}_{ ext{reaction}} = Σ( ext{moles of products} × ΔH_f^{ ext{°}}_{ ext{products}}) - Σ( ext{moles of reactants} × ΔH_f^{ ext{°}}_{ ext{reactants}})\]

This calculation was applied to the given reaction to find \( ΔH^{ ext{°}} = -55.99 ext{ kJ/mol} \), indicating an exothermic reaction, meaning it releases heat.

Equilibrium Constant

The Equilibrium Constant (\( K \)) is a vital parameter in chemical reactions, representing the ratio of concentrations of products to reactants at equilibrium. It provides a snapshot of the balance between products and reactants under specific conditions.

In a given chemical reaction, such as the one between NO, NO₂, and N₂O:

• If \( K \) is large, the reaction favors products at equilibrium.
• If \( K \) is small, reactants are favored.

The value of \( K \) can change with temperature, which is where the Van't Hoff Equation comes into play.

Van't Hoff Equation

The Van't Hoff Equation links the change in equilibrium constant (\( K \)) with temperature changes, based on the reaction's standard enthalpy change (\( ΔH^{ ext{°}} \)). The equation is as follows:

• \[\frac{d( ext{ln} \, K)}{dT} = \frac{ΔH^{ ext{°}}}{RT^2}\]

Where \( R \) is the gas constant and \( T \) is the temperature in Kelvin.

For reactions that are exothermic (\( ΔH^{ ext{°}} < 0 \)), as temperature increases, \( K \) decreases. This is intuitive, as adding heat to an exothermic reaction shifts the equilibrium towards the reactants. This principle helps in predicting how equilibrium compositions will be affected by changing temperatures.