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Chapter 7: Problem 11

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
Increase in temperature (option d) will shift the reaction forward.

Step by step solution

01

Understanding the Reaction and Enthalpy

The given reaction is \( \mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 21(\mathrm{~g}) \), and it is endothermic since \( \Delta \mathrm{H}^{\circ} = +150 \mathrm{~kJ} \). This means that the forward reaction absorbs heat.
02

Assessing the Options

We need to determine which changes will encourage the endothermic reaction to proceed in the forward direction. Consider the impact of each option on the equilibrium: (a) Increase in concentration of I, (b) Increase in total pressure, (c) Decrease in concentration of \( \mathrm{I}_{2} \), and (d) Increase in temperature.
03

Analyzing Option (a): Increase in Concentration of I

Increasing the concentration of I will shift the reaction towards the reactants (left) by Le Chatelier's Principle since more product is added.
04

Analyzing Option (b): Increase in Total Pressure

Increasing pressure shifts the equilibrium towards the side with fewer moles of gas. Here, the product side has more moles, so increasing pressure would favor the reactant side (left).
05

Analyzing Option (c): Decrease in Concentration of I2

Decreasing \( \mathrm{I}_{2} \) concentration shifts equilibrium towards the reactants (left) to make more \( \mathrm{I}_{2} \).
06

Analyzing Option (d): Increase in Temperature

Since the reaction is endothermic, an increase in temperature will favor the forward reaction (right), shifting the equilibrium towards the products to absorb the added heat.
07

Determining the Correct Answer

Among the given options, increasing temperature is the change that shifts the equilibrium in the forward direction, favoring the formation of more I due to the endothermic nature of the reaction.

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Key Concepts

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

Endothermic reactions
In chemistry, understanding endothermic reactions is crucial as they play an essential role in many natural and industrial processes. An endothermic reaction absorbs heat from the surroundings, which means it requires energy input to proceed. The positive \( \Delta \mathrm{H}^{\circ} \) value, such as \( +150\, \mathrm{kJ} \), indicates the amount of energy absorbed during the reaction.
Here are some key features of endothermic reactions to remember:
  • They absorb energy, which often makes the surroundings feel cooler.
  • The temperature increase can shift the equilibrium towards the products, as the system tries to balance by favoring the energy-consuming direction.
  • Common examples include photosynthesis and the melting of ice.
In our exercise, the reaction between iodine gas (\( \mathrm{I}_2(\mathrm{~g}) \)) and atomic iodine \( (2\mathrm{I}(\mathrm{~g})) \) is endothermic, supporting the notion that increasing temperature encourages the forward reaction.
Chemical equilibrium
Chemical equilibrium is a fascinating concept that describes a dynamic state in which the forward and reverse reactions occur at the same rate. At equilibrium, the concentrations of reactants and products remain constant over time.
Le Chatelier鈥檚 Principle helps explain how a system at equilibrium responds to changes in conditions:
  • If the concentration of reactants or products is altered, the equilibrium will shift to counteract the change and re-establish balance.
  • Changing the temperature can favor either the forward or reverse reaction, depending on whether the process is endothermic or exothermic.
  • Pressure changes affect equilibria involving gases, shifting towards the side with fewer moles if pressure is increased.
In our specific exercise, when temperature increases, it affects the equilibrium of the endothermic reaction \( \mathrm{I}_{2} \rightleftharpoons 2\mathrm{I} \), promoting the formation of more products. This aligns with the principle that endothermic reactions shift towards product formation with added heat.
Gaseous reactions
Gaseous reactions are a type of chemical reaction where the reactants, products, or both exist in the gas phase. These reactions are often influenced by changes in pressure and volume, in addition to concentration and temperature adjustments.
Some distinctive characteristics of gaseous reactions include:
  • They are highly responsive to pressure changes due to the gaseous nature of products and reactants.
  • The number of moles on each side of the equation determines how pressure changes influence equilibrium. More moles on one side means pressure increases favor that side.
  • Gaseous reactions often have faster reaction rates due to increased molecular movement and collision frequencies.
In our exercise, pressure increase favors the side with fewer gaseous moles, moving away from products towards reactants. This clarity helps understand how Le Chatelier's Principle is applied in gaseous systems like \( \mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2\mathrm{I}(\mathrm{~g}) \).

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Most popular questions from this chapter

In which of the following reactions, equilibrium is independent of pressure: (a) \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}) ; \Delta \mathrm{H}=+\mathrm{ve}\) (b) \(2 \mathrm{SO}_{2}+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g}) ; \Delta \mathrm{H}=-\mathrm{ve}\) (c) \(3 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{N}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) ; \Delta \mathrm{H}=-\mathrm{ve}\) (d) \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) ; \Delta \mathrm{H}=+\mathrm{ve}\)

One mole of HI was heated in a sealed tube at \(440^{\circ} \mathrm{C}\) till the equilibrium was reached. HI was found to be \(22 \%\) decomposed. The equilibrium constant for dissociation reaction, \(2 \mathrm{HI} \rightleftharpoons \mathrm{H}_{2}+\mathrm{I}_{2}\) is: (a) \(1.99\) (b) \(0.282\) (c) \(0.01988\) (d) \(0.0796\)

In which of the following reactions, the concentration of product is higher than the concentration of reactant at equilibrium? = equilibrium constant): (a) \(\mathrm{A} \rightleftharpoons \mathrm{B} ; \mathrm{K}=0.001\) (b) \(\mathrm{M} \rightleftharpoons \mathrm{N} ; \mathrm{K}=10\) (c) \(\mathrm{X} \rightleftharpoons \mathrm{Y} ; \mathrm{K}=0.005\) (d) \(\mathrm{R} \rightleftharpoons \mathrm{P} ; \mathrm{K}=0.01\)

\(1.25\) moles of NOCl were placed in a \(2.50 \mathrm{~L}\) reaction chamberat \(427^{\circ} \mathrm{C}\). After equilibrium was reached, 1.10 molesofNOClremained. Calculatetheequilibrium constant \(\mathrm{K}_{\mathrm{c}}\) for the reaction, \(2 \mathrm{NOC} 1(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}):\) (a) \(1.6 \times 10^{-3}\) (b) \(5.6 \times 10^{-4}\) (c) \(2.6 \times 10^{-3}\) (d) \(4.6 \times 10^{-4}\)

For which of the following reaction, \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}:\) (a) \(2 \mathrm{NOCl}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g})\) (b) \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) (c) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HCl}(\mathrm{g})\) (d) \(\mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{5}(\mathrm{~g})\)
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