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Which of the following change will shift the reaction in forward direction? \(\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons-21(\mathrm{~g})\) Take \(\Delta \mathrm{H}^{\circ}=+150 \mathrm{~kJ}\) (a) increase in concentration of \(I\) (b) increase in total pressure (c) decrease in concentration of \(\mathrm{I}_{2}\) (d) increase in temperature

Short Answer

Expert verified
Increasing the temperature (option d) will shift the reaction forward.

Step by step solution

01

Review Le Châtelier's Principle

Le Châtelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change and establish a new equilibrium.
02

Analyze the Effect of Concentration Changes

Changing the concentration of reactants or products will shift the equilibrium to oppose the change. Increasing the concentration of products will shift the reaction in the reverse direction, whereas decreasing it will shift it forward. Increased concentration of reactants will shift the equilibrium forward.
03

Evaluate the Impact of Concentration Changes

(a) Increasing the concentration of "I" is actually the opposite scenario because "I" is a product. Therefore, this would shift the equilibrium to the left. (c) Decreasing the concentration of "I2" would also shift the equilibrium to the right to produce more "I2" and counteract the reduction of "I2". Therefore, this would shift the equilibrium in the reverse direction.
04

Analyze the Effect of Pressure Changes

Changes in pressure affect reactions with different numbers of moles of gases on each side. If pressure is increased, equilibrium shifts towards the side with fewer moles of gas.
05

Evaluate the Impact of Increasing Total Pressure

(b) Increasing total pressure doesn't affect the equilibrium direction unless the number of moles of gases changes. Here, both sides of the equilibrium reaction have the same number of moles (one mole of "I2" decomposes to two moles of "I"), so there is no shift due to a pressure change.
06

Analyze the Effect of Temperature Changes

For endothermic reactions, where \(\Delta H^{\circ} = +150 \, \text{kJ}\), increasing the temperature shifts the equilibrium to the right, favoring the forward reaction.
07

Evaluate the Impact of Increasing Temperature

(d) Increasing the temperature provides additional energy that favors the endothermic direction, which in this case is the forward reaction, shifting the equilibrium to the right to produce more "I1".

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Key Concepts

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

Effects of Concentration Changes
Le Châtelier's Principle provides insight into how changing the concentration of reactants or products will affect a chemical equilibrium. In a balanced chemical reaction, altering the concentration of any of the components will cause the equilibrium to shift to counterbalance this change. For example:
  • If the concentration of a reactant is increased, the system will shift towards the products to consume the added reactant, thereby shifting the equilibrium forward.
  • Conversely, increasing the concentration of a product will shift the equilibrium towards the reactants, as the system tries to decrease the extra product.
  • Reducing the concentration of a reactant or adding more of a product will have the opposite effect, prompting the equilibrium to shift in the reverse direction.
Understanding these shifts is crucial for predicting how a system will respond to changes, and they play a significant role in chemical processes involving equilibrium.
Impact of Pressure Changes
Pressure changes can have a notable impact on chemical equilibria, particularly in reactions involving gases. According to Le Châtelier's Principle:
  • If a reaction involves gases and an increased pressure, the equilibrium will shift towards the side with fewer gas moles to alleviate the pressure change.
  • Conversely, decreasing the pressure will shift the equilibrium towards the side with more gas moles.
  • In reactions where the number of gas moles remains the same on both sides, pressure changes typically will not affect the equilibrium's position.
These principles help chemists understand and manipulate reactions by capitalizing on the pressure effects, particularly in industrial settings.
Temperature Effect on Equilibrium
Temperature changes affect equilibrium significantly, especially for reactions categorized as endothermic or exothermic. For an endothermic reaction, where heat is absorbed (indicated by \(\Delta H^{\circ} > 0\):
  • Raising the temperature supplies more heat to the system, effectively shifting the equilibrium towards the products since the reaction needs more heat, favoring progression in the forward direction.
  • If the temperature is decreased, the equilibrium shifts towards the reactants as the system attempts to generate more heat.
Temperature is often a controllable parameter in chemical reactions, allowing for the intentional shifting of equilibrium to produce desired products more efficiently.
Endothermic Reactions
Endothermic reactions absorb energy from their surroundings, usually in the form of heat, as indicated by a positive \(\Delta H^{\circ}\). In these reactions:
  • The system requires heat input to proceed in the forward direction, as the reactants turn into products through energy absorption.
  • Increasing the temperature adds energy, promoting the conversion of reactants into products and shifting the equilibrium towards the right (forward direction).
  • Conversely, lowering the temperature removes energy, causing a shift back towards the reactants as the system adjusts to absorb additional heat.
These principles are leveraged in various chemical and industrial processes to control product formation and yield.
Dynamic Equilibrium
A dynamic equilibrium occurs in reversible reactions when the rates of the forward and backward reactions are equal, resulting in no net change in the concentrations of reactants or products. Key characteristics include:
  • Both reactions continue to occur, but since their rates are equal, the observable concentrations remain constant over time.
  • The system is highly responsive to external changes such as concentration, pressure, and temperature, which can shift the equilibrium.
  • Despite these shifts, once the equilibrium is re-established, the forward and reverse reactions once again occur at the same rates.
Understanding dynamic equilibrium helps in predicting how a system can respond to external changes and maintain balance, a critical aspect in chemistry and various applications of reactions.

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Most popular questions from this chapter

For the reaction (1) and (2) \(\mathrm{A} \rightleftharpoons(\mathrm{g}) \quad \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})\) \(\mathrm{X} \rightleftharpoons(\mathrm{g}) \quad 2 \mathrm{Y}(\mathrm{g})\) Given, \(K_{p_{1}}: K_{p_{2}}=9: 1 .\) If degree of dissociation of \(A\) (g) and \(\mathrm{X}\) (g) are same then the ratio of total pressure in equilibrium (1) and (2) will be (a) \(36: 1\) (b) \(0.5: 1\) (c) \(1: 1\) (d) \(3: 1\)

Assertion: Adding an inert gas todissociation equilibrium of gaseous \(\mathrm{N}_{2} \mathrm{O}_{4}\) at constant pressure and temperature increases the dissociation. Reason: Molar concentration of the reactants and products decreases on the addition of inert gas.

A vessel at \(1000 \mathrm{~K}\) contains \(\mathrm{CO}_{2}\) with a pressure of \(0.5 \mathrm{~atm}\). Some of the \(\mathrm{CO}_{2}\) is converted into \(\mathrm{CO}\) on the addition of graphite. If the total pressure at equilibrium is \(0.8 \mathrm{~atm}\), the value of \(\mathrm{K}\) is: (a) \(3.6 \mathrm{~atm}\) (b) \(1 \mathrm{~atm}\) (c) \(2 \mathrm{~atm}\) (d) \(1.8 \mathrm{~atm}\)

\(\mathrm{K}_{\mathrm{p}}\) has the value of \(10^{-6} \mathrm{~atm}^{3}\) and \(10^{-4} \mathrm{~atm}^{3}\) at \(298 \mathrm{~K}\) and \(323 \mathrm{~K}\) respectively for the reaction \(\mathrm{CuSO}_{4} \cdot 3 \mathrm{H}_{2} \mathrm{O}(\mathrm{s}) \rightleftharpoons \mathrm{CuSO}_{4}(\mathrm{~s})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) \(\Delta_{r} \mathrm{H}^{\circ}\) for the reaction is: (a) \(7.7 \mathrm{~kJ} / \mathrm{mol}\) (b) \(-147.41 \mathrm{~kJ} / \mathrm{mol}\) (c) \(147.41 \mathrm{~kJ} / \mathrm{mol}\) (d) None of these

Match the following Column-I (a) \(\mathrm{N}_{2} \mathrm{O}_{4(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NO}_{2(g)}\) (b) \(\mathrm{PCl}_{s(\mathrm{~g})} \rightleftharpoons \mathrm{PCl}_{3(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})}\) (c) \(\mathrm{NH}_{2} \mathrm{COONH}_{4(\mathrm{~s})} \rightleftharpoons 2 \mathrm{NH}_{3(\mathrm{~g})}+\mathrm{CO}_{2 \mathrm{~g}}\) (d) \(\mathrm{H}_{2(g)}+\mathrm{I}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{HI}_{(\mathrm{g})}\) Column-II (p) \(K_{P}=\frac{4 x^{2} p}{a^{2}-x^{2}}\) (q) \(K_{P}=\frac{4 P^{3}}{27}\) (r) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}\) (s) \(K_{p}=K_{c^{\circ}} R T\) (t) \(\Delta \mathrm{n}_{\mathrm{g}}>0\)

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