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For the system at equilibrium which of the following are correct? (a) On increasing the temperature of an endothermic reaction, the equilibrium shifts in forward direction because Q decreases. (b) On increasing the temperature of an endothermic reaction, the concentration in moles per litre of the reactants increases. (c) \(\log \mathrm{K}=\frac{1}{2.303 \mathrm{R}}\left(\Delta \mathrm{S}^{\circ}-\frac{\Delta \mathrm{H}^{\circ}}{\mathrm{T}}\right)\) (d) On increasing the temperature of an endothermic reaction, the equilibrium shifts in forward direction because \(\mathrm{K}\) increases.

Short Answer

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(c) and (d) are correct.

Step by step solution

01

Understand the Context

The exercise deals with the effect of temperature changes on the equilibrium of an endothermic reaction. Endothermic reactions absorb heat, hence adding more heat (increasing temperature) affects the equilibrium position based on Le Chatelier's Principle.
02

Analyze Option (a)

For an endothermic reaction, increasing temperature generally leads to the forward reaction being favored as it absorbs energy. However, \(Q\) (reaction quotient) itself doesn't necessarily decrease; rather, what changes is the equilibrium constant \(K\). Thus, this statement is not conceptually correct.
03

Analyze Option (b)

For an endothermic reaction, increasing temperature should shift the equilibrium towards the products, leading to a decrease in the reactants' concentration in moles per litre, not an increase. Hence, this option is incorrect.
04

Analyze Option (c)

The given equation \(\log K = \frac{1}{2.303 R}\left(\Delta S^{\circ} - \frac{\Delta H^{\circ}}{T}\right)\) is related to the van 't Hoff equation which describes how an equilibrium constant \(K\) changes with temperature. The equation is correct.
05

Analyze Option (d)

In an endothermic reaction, increasing the temperature usually results in an increase in the equilibrium constant \(K\), favoring the formation of products (forward reaction). Therefore, this statement correctly describes the effect of increasing temperature on the equilibrium position.

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Key Concepts

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

Le Chatelier's Principle
Le Chatelier's Principle is a foundational concept in chemistry that helps us predict how a change in a system at equilibrium will affect that system. Simply put, it states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change.
  • If a reactant or product is added to a system, the system will shift to consume the added substance.
  • If the temperature is increased in an endothermic reaction, the equilibrium will shift to absorb the extra heat.
  • If the system is compressed by increasing pressure, the equilibrium will shift towards the side with fewer moles of gas.
Le Chatelier's Principle is useful in predicting how the concentrations of reactants and products will change in response to various kinds of stress. It gives us a way to visualize the process and make educated guesses about how the system behaves under new conditions.
Endothermic Reactions
Endothermic reactions are chemical reactions that absorb energy from their surroundings, primarily in the form of heat. This is an important aspect when discussing chemical equilibrium and how to manipulate it.
  • In these reactions, heat can be considered a reactant. Adding heat shifts the equilibrium towards the products.
  • An increase in temperature typically increases the rate of the reaction, hence more products form.
  • The positive change in enthalpy (\( \Delta H \)) characterizes these reactions.
Understand that endothermic reactions are crucial for various industrial and chemical processes. For instance, the thermal decomposition of calcium carbonate in the production of lime involves endothermic processes that are temperature-dependent.
Temperature Effect on Equilibrium
Temperature is a key factor that can pivot the equilibrium of a reaction. According to Le Chatelier’s Principle, changing the temperature of a system in equilibrium will cause the system to readjust to regain equilibrium. The effect of temperature on an equilibrium directly depends on whether the reaction is exothermic or endothermic.
  • In endothermic reactions, an increase in temperature shifts equilibrium towards products.
  • Conversely, in exothermic reactions, decreasing the temperature also shifts equilibrium towards the products.
  • This effect can be described quantitatively using the Van 't Hoff Equation.
Understanding how temperature affects equilibrium can be crucial for optimizing conditions in various chemical processes, ensuring maximal yield and efficiency.
Van 't Hoff Equation
The Van 't Hoff Equation gives us a powerful tool for understanding the relationship between the temperature and the equilibrium constant (\( K \)) of a reaction. This equation helps predict how \( K \) changes with temperature, a vital factor in chemical kinetics and thermodynamics.
The equation is as follows:
\[ \log K = \frac{1}{2.303 R}\left( \Delta S^{\circ} - \frac{\Delta H^{\circ}}{T} \right) \]
Where:
  • \( K \) is the equilibrium constant
  • \( \Delta S^{\circ} \) is the standard entropy change
  • \( \Delta H^{\circ} \) is the standard enthalpy change
  • \( T \) is the temperature in Kelvin
  • \( R \) is the universal gas constant
The Van 't Hoff Equation is beneficial when evaluating reactions that are sensitive to temperature changes, aiding chemists in controlling and predicting the outcomes of reactions.

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Most popular questions from this chapter

Considering entropy (S) as a thermodynamic parameter, the criterion for the spontaneity of any process is (a) \(\Delta \mathrm{S}_{\text {system }}+\Delta \mathrm{S}_{\text {surroundings }}>0\) (b) \(\Delta \mathrm{S}_{\text {system }}-\Delta \mathrm{S}_{\text {surroundings }}>0\) (c) \(\Delta \mathrm{S}_{\text {system }}>0\) (d) \(\Delta \mathrm{S}_{\text {surroundings }}>0\)

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Which of the following is /are true about the isothermal expansion of an ideal gas? (a) \(\Delta \mathrm{U}=0\) (b) \(\Delta \mathrm{T}=0\) (c) \(\mathrm{q}=2.303 \mathrm{nRT} \log _{10}\left(\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}\right)\) (d) \(\mathrm{q}=0\)

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