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

The standard enthalpy change (∆H°) for the reaction is -65.3 kJ/mol. Since the reaction is exothermic, the equilibrium constant (K) will decrease with increasing temperature. At constant temperature, increasing the volume will favor the reactants, increasing the fraction of products (NO₂ and N₂O) in the equilibrium mixture. Conversely, decreasing the volume will favor the products, decreasing the fraction of products in the equilibrium mixture.

Step-by-step solution

Find the standard enthalpies of formation

Firstly, we need to find the standard enthalpies of formation (∆H°f) for each gas from Appendix C. The values are as follows: \(∆H°_{f} (NO(g)) = 90.29 \, kJ/mol\) \(∆H°_{f} (NO_{2}(g)) = 33.83 \, kJ/mol\) \(∆H°_{f} (N_{2}O(g)) = 81.57 \, kJ/mol\)

Calculate ∆H° for the reaction

Using the standard enthalpies of formation, we can calculate the enthalpy change for the reaction as follows: ∆H° = [ ∆H°f(products) - ∆H°f(reactants) ] ∆H° = [ \(1 \times 33.83 + 1 \times 81.57\) ] - [ \(3 \times 90.29\) ] ∆H° = -65.3 kJ/mol

Determine the effect of temperature on equilibrium constant

Next, we need to understand the relationship between temperature and the equilibrium constant. According to Le Chatelier's principle, if the reaction is exothermic (i.e., ∆H° < 0), increasing the temperature will shift the equilibrium towards the side with fewer moles of gas (favoring the reactants in this case). Conversely, if the reaction is endothermic (i.e., ∆H° > 0), increasing the temperature will shift the equilibrium towards the side with more moles of gas (favoring the products in this case). In our case, since ∆H° is negative (-65.3 kJ/mol), the reaction is exothermic, and the equilibrium will shift towards the reactants when the temperature is increased. Therefore, the equilibrium constant (K) will decrease with increasing temperature.

Determine the effect of volume change on equilibrium mixture

Lastly, we need to analyze the effect of volume change on the equilibrium mixture at constant temperature. According to Le Chatelier's principle, a decrease in volume increases the pressure, which will favor the side with fewer moles of gas, while an increase in volume reduces the pressure, favoring the side with more moles of gas. In our reaction, the number of moles on the reactant side (3 moles of NO) is greater than the number of moles on the product side (1 mole of NO₂ and 1 mole of N₂O). Thus, at constant temperature, an increase in the volume of the container will shift the equilibrium towards the reactants, increasing the fraction of products in the equilibrium mixture (NO₂ and N₂O). Conversely, a decrease in the container's volume will shift the equilibrium towards the products, decreasing the fraction of products in the equilibrium mixture.

Key concepts

Le Chatelier's Principle

Le Chatelier's Principle is a cornerstone concept in chemical equilibrium that helps predict how a system at equilibrium reacts to external changes. Imagine a seesaw perfectly balanced at equilibrium. If we add weight to one side, the seesaw will tilt to restore balance. Similarly, a chemical system at equilibrium will adjust to counteract the imposition of changes such as alterations in concentration, pressure, or temperature.

For example, if we increase the concentration of a reactant, the system shifts to produce more products to reduce the reactants' heightened concentration. Likewise, for pressure, which is akin to crowding, if we reduce the container size (increasing the pressure), the system shifts toward the side with fewer gas particles, essentially seeking more 'elbow room'. Temperature changes are analogous to adding or removing thermal energy: when we heat an equilibrium mixture of an exothermic reaction (releases heat), the system adjusts to absorb excess heat by favoring the reactant side. In the case of our nitrogen oxides reaction, the decrease in temperature would therefore shift the equilibrium toward product formation, conversely to the increase in temperature.

Enthalpy Change Calculation

In thermodynamics, the enthalpy change (abla H) of a reaction signifies the total heat absorbed or released under constant pressure. It’s a measure of the energy change in a system and plays a vital role in determining whether a reaction is endothermic (absorbing heat) or exothermic (releasing heat). To calculate the enthalpy change, you apply the following formula:
abla H = [ abla H^°f(products) - abla H^°f(reactants) ].

This formula hinges on the standard enthalpies of formation (abla H°f), which are the heat changes when one mole of a compound forms from its elements at standard conditions. For reactions including our equilibrium between oxides of nitrogen, we look up these abla H°f values from a reference like Appendix C, then subtract the total abla H°f of the reactants from that of the products. An essential tip to simplify calculations is to remember for elements in their standard states, such as N₂(g), abla H°f is zero. If the result is negative, it signals the reaction is exothermic, releasing heat and likely to decrease in rate with rising temperature.

Equilibrium Constant

The equilibrium constant (K) is a quantitative measure of the proportion of reactants to products at equilibrium. It's derived from the law of mass action, which in its core states the rate of a reaction is proportional to the concentration of the reactants raised to a power equivalent to their coefficients in the balanced equation. For the generalized reaction aA + bB ↔ cC + dD, the equilibrium constant expression is
K = [C]^c [D]^d / [A]^a [B]^b.

The beauty of the equilibrium constant is its invariant nature at a fixed temperature. No matter the starting concentrations, the system will adjust until K is reached. In our nitrogen oxides example, if you changed the temperature, you'd effectively alter the value of K, since it is temperature-dependent. An exothermic reaction (with a negative abla H) will have a K that decreases with increasing temperature, as the system favors the reactant side to dissipate added heat. In contrast, for endothermic reactions, K increases as temperature rises. Understanding K helps in predicting the concentration of reactants and products without knowing the specific dynamics at play.