Question

Consider the following reactions. In which cases is the product formation favoured by decreased pressure? (1) \(\mathrm{CO}_{2}(g)+\mathrm{C}(s) \rightleftharpoons 2 \mathrm{CO}(g) ; \quad \Delta H^{\circ}=+172.5 \mathrm{~kJ}\) (2) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) ; \quad \Delta H^{\circ}=-91.8 \mathrm{~kJ}\) (3) \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) ; \quad \Delta H^{\circ}=181 \mathrm{~kJ}\) (4) \(2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) ; \quad \Delta H^{\circ}=484.6 \mathrm{~kJ}\) (a) 2,3 (b) 3,4 (c) 2,4 (d) 1,4

Short answer

(d) 1,4

Step-by-step solution

Consider Le Chatelier's Principle

Le Chatelier's Principle states that if a system at equilibrium is subjected to a change in concentration, temperature, volume, or pressure, then the system adjusts its equilibrium to counteract the effect of the applied change. Decreasing the pressure of a gas reaction shifts the point of equilibrium in favor of the side with more moles of gas, since the volume available to the gas molecules increases.

Analyze Reaction (1)

The reaction CO2(g) + C(s) rightleftharpoons 2 CO(g) shows an increase in the number of moles of gas from one mole on the left to two moles on the right. Decreasing pressure would shift the equilibrium to the right, favoring the formation of the products.

Analyze Reaction (2)

In the reaction N2(g) + 3 H2(g) rightleftharpoons 2 NH3(g), there are 4 moles of gas on the left and 2 moles on the right. Decreasing pressure would shift the equilibrium to the left, favoring the reactants rather than the formation of products.

Analyze Reaction (3)

The reaction N2(g) + O2(g) rightleftharpoons 2 NO(g) starts with 2 moles of gases and forms 2 moles of gases. A change in pressure will not favor either the reactants or the products as the number of moles of gas remains the same on both sides of the equilibrium.

Analyze Reaction (4)

In the reaction 2 H2O(g) rightleftharpoons 2 H2(g) + O2(g), there are 2 moles of gas on the left and 3 moles on the right. Decreasing the pressure would shift the equilibrium to the right, favoring the formation of the products.

Choose the Correct Answer

Based on the analysis, we can see that decreasing pressure favors the product formation in Reaction (1) and Reaction (4). Therefore, the correct answer is (d) 1,4.

Key concepts

Chemical Equilibrium

Understanding chemical equilibrium is essential for grasping many concepts in chemistry, including reaction dynamics and how different factors can shift equilibrium. Equilibrium occurs when the forward and reverse reactions happen at the same rate, resulting in no net change in the concentration of the reactants and products over time. At this point, the system is said to be balanced, but this does not mean the reactants and products are present in equal amounts—rather, their ratios are constant.

For a reaction such as \( A \rightleftharpoons B \), where one mole of A turns into one mole of B, at equilibrium both A and B will be produced and used up at the same rate. The equilibrium constant, represented as \( K_{eq} \), becomes an invaluable tool for chemists to understand the extent of reactions and predict how changes in conditions might affect the reaction. It is calculated by taking the ratio of the concentration of the products raised to their stoichiometric coefficients over the concentration of the reactants raised to their coefficients.

Equilibrium Shifts and Pressure

According to Le Chatelier's Principle, if an external change is imposed on a reaction system at equilibrium, the system will adjust to minimize that change. When it comes to gases, pressure becomes a critical external factor. A change in pressure can be accomplished by changing the volume of the gas or directly adding or removing a gas.

In a contained system with gases, decreasing the volume will increase pressure and drive the equilibrium position toward the side with fewer moles of gas, to decrease the pressure. Conversely, if the pressure is decreased by increasing the volume, the equilibrium will shift towards the side with more moles of gas. This is beautifully illustrated by envisioning gas molecules having more space to move around when pressure is decreased (volume is increased), favoring the side of the reaction that produces more gas molecules.

Enthalpy Changes in Reactions

Enthalpy change, denoted as \( \Delta H \), in reactions refers to the heat released or absorbed under constant pressure. It is an essential concept as it indicates whether a reaction is exothermic (releases heat, \( \Delta H < 0 \)) or endothermic (absorbs heat, \( \Delta H > 0 \)). This change in heat can also affect chemical equilibrium.

An exothermic reaction, upon experiencing a temperature increase, will be shifted towards the reactants to absorb the excess heat (Le Chatelier's Principle). On the other hand, an endothermic reaction will shift towards producing more products when the temperature is raised, because it consumes heat. Hence, understanding \( \Delta H \) allows chemists to predict and control the direction of equilibrium shifts due to temperature changes.

Mole Concept in Gases

The mole concept is fundamental in chemistry, especially when dealing with gases. A mole represents Avogadro's number (approximately \(6.022 \times 10^{23}\)) of particles, be they atoms, molecules, ions, or electrons. In terms of gases, Avogadro's law states that at the same temperature and pressure, equal volumes of all gases contain the same number of moles.

This concept allows for predictable behavior of gases under changing conditions and is used to understand gas reactions in equilibrium. For instance, if a reaction produces more moles of gas, the overall volume of gas will increase if the temperature and pressure are kept constant, explaining why equilibrium shifts towards more moles of gas when pressure is decreased. Understanding the interplay between moles, volume, and pressure is vital for anyone studying gas equilibria.