Which of the following reactions will get affected by increase of pressure? Also mention, whether change will cause the reaction to go into the right or left direction? (i) \(\mathrm{CH}_{4(\mathrm{~g})}+2 \mathrm{~S}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{CS}_{2(\mathrm{~g})}+2 \mathrm{H}_{2} \mathrm{~S}_{(\mathrm{g})}\) (ii) \(\mathrm{CO}_{2(\mathrm{~g})}+\mathrm{C}_{(\mathrm{s})} \rightleftharpoons 2 \mathrm{CO}_{(\mathrm{g})}\) (iii) \(4 \mathrm{NH}_{3(\mathrm{~g})}+5 \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons 4 \mathrm{NO}_{(\mathrm{g})}+6 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})}\) (iv) \(\mathrm{C}_{2} \mathrm{H}_{4(\mathrm{~g})}+\mathrm{H}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{6(\mathrm{~g})}\)

Short Answer

Expert verified
The reactions (ii), (iii), and (iv) will be affected by an increase in pressure. The increased pressure will cause reaction (ii) to shift left, favoring reactants; reaction (iii) to shift left, favoring reactants; and reaction (iv) to shift right, favoring product.

Step by step solution

01

Understanding Le Chatelier's Principle

Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. In terms of pressure, this means that if the pressure is increased, the equilibrium will shift towards the side with fewer moles of gas to reduce the pressure. Conversely, if pressure is decreased, the equilibrium will shift towards the side with more moles of gas.
02

Analyze Reaction (i)

For the reaction \( \mathrm{CH}_{4(\mathrm{g})} + 2 \mathrm{S}_{2(\mathrm{g})} \rightleftharpoons \mathrm{CS}_{2(\mathrm{g})} + 2 \mathrm{H}_{2} \mathrm{S}_{(\mathrm{g})} \), there are 3 moles of gaseous reactants and 3 moles of gaseous products. Since the number of gas moles on both sides is equal, increasing the pressure will have no effect on the position of equilibrium for this reaction.
03

Analyze Reaction (ii)

For the reaction \( \mathrm{CO}_{2(\mathrm{g})} + \mathrm{C}_{(\mathrm{s})} \rightleftharpoons 2 \mathrm{CO}_{(\mathrm{g})} \), there is 1 mole of gaseous reactant and 2 moles of gaseous products. Increasing pressure will shift the equilibrium to the left towards the side with fewer gas moles, which is the reactant side. Thus, the reaction will favor the formation of \( \mathrm{CO}_{2} \) and \( \mathrm{C} \).
04

Analyze Reaction (iii)

For the reaction \( 4 \mathrm{NH}_{3(\mathrm{~g})} + 5 \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons 4 \mathrm{NO}_{(\mathrm{g})} + 6 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})} \), there are 9 moles of gaseous reactants and 10 moles of gaseous products. Increasing the pressure will shift the equilibrium to the left, favoring the formation of reactants, \( \mathrm{NH}_{3} \) and \( \mathrm{O}_{2} \).
05

Analyze Reaction (iv)

For the reaction \( \mathrm{C}_{2} \mathrm{H}_{4(\mathrm{~g})} + \mathrm{H}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{6(\mathrm{~g})} \), there are 2 moles of gaseous reactants and 1 mole of gaseous product. Increasing the pressure will shift the equilibrium to the right, favoring the formation of \( \mathrm{C}_{2} \mathrm{H}_{6} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equilibrium
Chemical equilibrium occurs in a closed system when the rates of the forward and reverse reactions are equal, leading to constant concentrations of the reactants and products. It is a dynamic state, which means that while the concentrations remain unchanged, the reactions continue to occur at a molecular level. Think of it as a tug-of-war where both teams are equally strong; the rope doesn’t move, but both teams are still pulling.

Within this balanced state, however, equilibrium can be disturbed by changing the temperature, pressure, or concentration of reactants or products. This is where Le Chatelier's Principle comes into play, which predicts how the system will respond to relieve the stress caused by these changes, and works to restore a new state of equilibrium.
Equilibrium Shift and Pressure
When dealing with gas phase reactions, pressure changes can significantly influence chemical equilibrium. According to Le Chatelier's Principle, an increase in pressure will cause the equilibrium to shift towards the side with fewer moles of gas. This is because the system aims to decrease the pressure by favoring the production of fewer gas molecules.

On the other hand, reducing pressure will shift the equilibrium toward the side with more moles of gas. This is the system’s way of filling the expanded volume. These shifts are pertinent to processes where gases are involved and have practical implications in industrial chemical reactions where control over product yield is essential.
Gas Phase Reactions
Gas phase reactions are those that occur within the gaseous state. These types of reactions are susceptible to changes in pressure, as gases are compressible and their molecules can move closer together or farther apart when pressure changes. This property directly relates to how equilibrium shifts occur in response to pressure changes, as proposed by Le Chatelier’s Principle.

In the gas phase, the number of moles of reactants and products becomes a pivotal factor in predicting the direction of the shift. This forms the basis for understanding the effect of pressure on equilibrium in gases and highlights why reactions involving gases are often studied concerning their response to pressure variations.
Moles and Gaseous Reactants
The concept of moles is fundamental to understanding chemical reactions and equilibrium, especially in the gas phase. A mole is a unit that measures the amount of substance, based on the number of particles, such as atoms or molecules, it contains. In the context of gaseous reactants, counting moles is crucial. It helps determine the amounts of reactants and products at equilibrium.

Le Chatelier’s Principle relies on mole counts to forecast how equilibrium will shift in response to pressure changes. If a reaction side has more moles of gas, it represents a larger volume and pressure. Therefore, a change in system pressure can shift equilibrium towards the side with fewer gas moles to minimize the impact of that change. This relationship underscores the importance of moles in predicting the behavior of gaseous reactants under varying conditions.

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Most popular questions from this chapter

Calculate the concentration of \(\mathrm{CO}_{2}\) in a soft drink that is bottled with a partial pressure of \(\mathrm{CO}_{2}\) of \(4.0 \mathrm{~atm}\) over the liquid at \(25^{\circ} \mathrm{C}\). The Henry's law constant for \(\mathrm{CO}_{2}\) in water is \(3.1 \times 10^{-2} \mathrm{~mol} /\) litre-atm at this temperature.

\(100 \mathrm{~g}\) of \(\mathrm{NaCl}\) is stirred in \(100 \mathrm{~mL}\) of water at \(20^{\circ} \mathrm{C}\) till the equilibrium is attained : (a) How much \(\mathrm{NaCl}\) goes into the solution and how much of it is left undissolved at equilibrium? The solubility of \(\mathrm{NaCl}\) at \(20^{\circ} \mathrm{C}\) is \(6.15\) mol/litre. (b) What will be the amount of \(\mathrm{NaCl}\) left undissolved if the solution is diluted to \(200 \mathrm{~mL} ?\)

If a mixture of 3 moles of \(\mathrm{H}_{2}\) and one mole of \(\mathrm{N}_{2}\) is completely converted into \(\mathrm{NH}_{3}\). What would be the ratio of the initial and final volume at same temperature and pressure?

A compound formed by elements \(A\) and \(B\) crystallizes in cubic structure where \(A\) atoms are at the corners of a cube and \(B\) atoms are at the face centre. What is the formula of the compound?

For the gaseous reaction; \(2 \mathrm{NO}_{2} \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}\), calculate \(\Delta \mathrm{G}^{\circ}\) and \(K_{\mathrm{p}}\) for the reaction at \(25^{\circ} \mathrm{C}\). Given \(\mathrm{G}_{j}^{\circ} \mathrm{N}_{2} \mathrm{O}_{4}\) and \(\mathrm{G}_{f}^{\circ} \mathrm{NO}_{2}\) are \(97.82\) and \(51.30\) \(\mathrm{kJ}\) respectively. Also calculate \(\Delta \mathrm{G}^{\circ}\) and \(K_{\mathrm{p}}\) for reverse reaction.

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