Question

Product the shift in equilibrium when the volume is decreased on the following equilibrium reactions: (i) \(\mathrm{Pcl}_{3(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})} \rightleftarrows \mathrm{PCl}_{5(\mathrm{~g})}\) (ii) \(\mathrm{N}_{2(g)}+\mathrm{O}_{2(\mathrm{~g})} \rightleftarrows 2 \mathrm{NO}_{(g)}\)

Short answer

Answer: In the first reaction, the equilibrium will shift to the right (towards the products), forming more PCl5(g). In the second reaction, there will be no significant effect on the equilibrium position as both sides have equal moles of gas.

Step-by-step solution

Identify the number of moles of reactants and products for each reaction

To understand how the equilibrium will shift when volume is decreased, it's essential to identify the ratio of moles of reactants/products for each reaction. (i) In the first reaction: \(\mathrm{Pcl}_{3(g)}+\mathrm{Cl}_{2(g)} \rightleftarrows \mathrm{PCl}_{5(g)}\) There is 1 mole of \(\mathrm{Pcl}_{3(g)}\) and 1 mole of \(\mathrm{Cl}_{2(g)}\) (total of 2 moles) on the left (reactants) side, and 1 mole of \(\mathrm{PCl}_{5(g)}\) on the right (product) side. (ii) In the second reaction: \(\mathrm{N}_{2(g)}+\mathrm{O}_{2(g)} \rightleftarrows 2\mathrm{NO}_{(g)}\) There is 1 mole of \(\mathrm{N}_{2(g)}\) and 1 mole of \(\mathrm{O}_{2(g)}\) (total of 2 moles on the left (reactants) side, and 2 moles of \(\mathrm{NO}_{(g)}\) on the right (product) side.

Applying Le Chatelier's principle

To predict how the equilibrium will shift upon a decrease in volume, we need to apply Le Chatelier's principle. Decreasing the volume of the container increases the pressure inside the container. (i) For the first reaction, we have more moles of gas on the reactant (left) side. Therefore, when the pressure increases due to a decrease in volume, the system will try to counteract the change by shifting the equilibrium to the side with fewer moles of gas. In this case, the equilibrium will shift to the right (towards the products) to minimize the pressure increase, forming more \(\mathrm{PCl}_{5(g)}\). (ii) For the second reaction, there are equal moles of gas on both sides of the equation (2 moles each). Therefore, when the volume is decreased, and the pressure increases, there will be no significant effect on the equilibrium position. This is because the system has equal moles on both sides, and shifting the equilibrium will not help to counteract the change in pressure.

Key concepts

chemical equilibrium

Chemical equilibrium refers to the state in a reversible reaction where the rate of the forward reaction equals the rate of the backward reaction. This state results in constant concentrations of reactants and products over time.
Even though the concentrations of the substances involved are stable, it doesn’t mean they are equal; it just means they no longer change.

• Equilibrium can be dynamic, as reactions continue to occur, but there is no net change in the composition of the system.
• At equilibrium, the ratio of the products to reactants is described by the equilibrium constant, denoted as K, which is unique for every reaction at a given temperature.

Understanding chemical equilibrium is essential for predicting how changes in the conditions can affect the system, a primary focus of Le Chatelier's Principle.

effect of pressure on equilibrium

The effect of pressure changes on chemical equilibrium is an important consideration in gaseous reactions. According to Le Chatelier's Principle, if a system at equilibrium is disturbed, the system will adjust to counteract the change and restore a new equilibrium state.

• In a gaseous system, decreasing the volume increases the pressure.
• The system responds by shifting the equilibrium towards the side with fewer moles of gas, which helps to decrease the pressure.

For instance, in the reaction i. \(\mathrm{Pcl}_{3(g)}+\mathrm{Cl}_{2(g)} \rightleftarrows \mathrm{PCl}_{5(g)}\): - There are more moles on the left (reactants) side. When the volume decreases, the equilibrium shifts to the right, towards fewer moles on the product side. However, if the moles of gas do not differ on both sides of the reaction, as in ii. \(\mathrm{N}_{2(g)}+\mathrm{O}_{2(g)} \rightleftarrows 2\mathrm{NO}_{(g)}\): - There is no shift, since moles are equal on both sides, and no pressure imbalance to correct.

mole concept

The mole concept is vital for calculating the amounts of reactants and products in chemical reactions. A mole is a unit used to express amounts of a chemical substance and contains Avogadro's number of particles or entities, which is \(6.022 \times 10^{23}\).
Understanding how to calculate and use moles helps in predicting the outcomes of chemical reactions under varying conditions.

• Moles are used to balance chemical equations and to calculate the reagents and products involved.
• In equilibrium reactions, identifying the number of moles on each side of the equation assists in predicting how the reaction will respond to changes.

In the given exercise, the mole concept helps recognize the significant difference in moles on the reactant and product sides for different reactions, thus aiding in the understanding of equilibrium shifts due to pressure changes.

reaction shift prediction

Predicting how reactions will shift in response to changes is a key aspect of chemistry that helps in controlling industrial chemical processes, optimizing yields, and understanding natural phenomena.

• Le Chatelier's Principle is used extensively in predicting reaction shifts—it states that a system in equilibrium will counteract applied stresses to regain equilibrium.
• When pressure increases (due to decreased volume), shifts depend on the number of moles of gas on each side of the reaction equation.

In reaction i. \(\mathrm{Pcl}_{3(g)}+\mathrm{Cl}_{2(g)} \rightleftarrows \mathrm{PCl}_{5(g)}\), - The reaction shifts right towards products because there are fewer moles of gas on the product side. For reaction ii. \(\mathrm{N}_{2(g)}+\mathrm{O}_{2(g)} \rightleftarrows 2\mathrm{NO}_{(g)}\), - The equilibrium does not shift because the gas moles are balanced on both sides. Accurate predictions allow chemists to manipulate conditions to achieve desired reaction outcomes.