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?

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.

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.

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.