
In the gaseous equilibrium ${{H}_{2}}{{X}_{2}}+heat\rightleftharpoons 2HX$,the formation of $HX$will be favored by [CPMT$1977$]
A. High pressure and low temperature
B. High temperature and low pressure
C. Low temperature and low pressure
D. High temperature and high pressure
Answer
164.1k+ views
Hint: The compound ${{H}_{2}}{{X}_{2}}$absorbs heat to form $HX$. This is an example of an endothermic reaction. So, the system will try to restore equilibrium as the temperature of the system decreases and also the number of moles increases in the gaseous equilibrium in such a way as to neutralize the changes produced at equilibrium.
Complete answer:According to Le-Chatelier’s principle, if any system is disturbed by pressure, concentration, or temperature, then the position of equilibrium shifts to neutralize the change to reestablish a new equilibrium.
In the gaseous equilibrium, ${{H}_{2}}{{X}_{2}}+heat\rightleftharpoons 2HX$
This equilibrium prefers endothermic reaction as it takes energy. For endothermic reactions, according to Le-Chatelier’s principle, an increase in temperature will enhance product formation and a decrease in temperature will retard product formation. So, the formation $HX$will be favored by high temperatures.
As the number of moles increases in the forward direction because here one mole ${{H}_{2}}{{X}_{2}}$forms two moles $HX$. In this condition according to Le-Chatelier’s principle when the number of moles increases, the equilibrium shifts in the direction in which there is a decrease in pressure.
When the number of moles increases the volume of the system also increases. Then the pressure of the system decreases as we know Volume is inversely proportional to pressure.
Therefore the formation $HX$will be favored by high temperature and low pressure.
Thus, option (B) is correct.
Note: The scenario is different in the case of an exothermic reaction. According to Le-Chatelier’s principle, an increase in temperature will decrease product formation and a decrease in temperature will increase product formation.
Complete answer:According to Le-Chatelier’s principle, if any system is disturbed by pressure, concentration, or temperature, then the position of equilibrium shifts to neutralize the change to reestablish a new equilibrium.
In the gaseous equilibrium, ${{H}_{2}}{{X}_{2}}+heat\rightleftharpoons 2HX$
This equilibrium prefers endothermic reaction as it takes energy. For endothermic reactions, according to Le-Chatelier’s principle, an increase in temperature will enhance product formation and a decrease in temperature will retard product formation. So, the formation $HX$will be favored by high temperatures.
As the number of moles increases in the forward direction because here one mole ${{H}_{2}}{{X}_{2}}$forms two moles $HX$. In this condition according to Le-Chatelier’s principle when the number of moles increases, the equilibrium shifts in the direction in which there is a decrease in pressure.
When the number of moles increases the volume of the system also increases. Then the pressure of the system decreases as we know Volume is inversely proportional to pressure.
Therefore the formation $HX$will be favored by high temperature and low pressure.
Thus, option (B) is correct.
Note: The scenario is different in the case of an exothermic reaction. According to Le-Chatelier’s principle, an increase in temperature will decrease product formation and a decrease in temperature will increase product formation.
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