
${{N}_{2}}+{{O}_{2}}\rightleftharpoons 2NO-Q$Cals
In the above reaction which is the essential condition for the higher production of $NO$
A. High temperature
B. High pressure
C. Low temperature
D. Low pressure
Answer
233.1k+ views
Hint: When any system in equilibrium is subject to a change in pressure, concentration, and temperature either the system changes to a new equilibrium or this change partially counteracts the applied change. When the temperature at equilibrium increases the system will shift toward the reactant side where the temperature of the reaction decreases.
Complete answer:According to Le-Chatelier’s principle, when temperature increases the position of equilibrium will shift in such a way as to reduce the temperature of the system.
In exothermic reactions energy in the form of heat is released toward the product side whereas in an endothermic reaction energy is absorbed towards the reactant side or uses more energy than released. In this case, the temperature can be considered as a reactant and product of a reaction as the temperature is a measure of the energy of the system. In an exothermic reaction, the temperature is considered to be a product and in an endothermic reaction, the temperature is considered to be a reactant.
$\text{Reactant}\to \text{Product+Heat energy}$ [In exothermic reaction]
$\operatorname{Re}ac\tan t\to \Pr oduct-Heat$energy [In endothermic reaction]
Here in this question, $Q$Calorie heat is absorbed by the reactant than released or the system uses more energy of $Q$calories than released, which means it is an endothermic reaction. Therefore for higher production $NO$, the system will shift towards the side where the temperature of the system increases.
From the given reaction higher temperature is the essential condition for higher production
$NO$.
Thus, option (A) is correct.
Note: In addition temperature, concentration and pressure change also affect the overall system If pressure or concentration increases at equilibrium the reaction will shift towards the direction with lower pressure or lower concentration side and vice versa.
Complete answer:According to Le-Chatelier’s principle, when temperature increases the position of equilibrium will shift in such a way as to reduce the temperature of the system.
In exothermic reactions energy in the form of heat is released toward the product side whereas in an endothermic reaction energy is absorbed towards the reactant side or uses more energy than released. In this case, the temperature can be considered as a reactant and product of a reaction as the temperature is a measure of the energy of the system. In an exothermic reaction, the temperature is considered to be a product and in an endothermic reaction, the temperature is considered to be a reactant.
$\text{Reactant}\to \text{Product+Heat energy}$ [In exothermic reaction]
$\operatorname{Re}ac\tan t\to \Pr oduct-Heat$energy [In endothermic reaction]
Here in this question, $Q$Calorie heat is absorbed by the reactant than released or the system uses more energy of $Q$calories than released, which means it is an endothermic reaction. Therefore for higher production $NO$, the system will shift towards the side where the temperature of the system increases.
From the given reaction higher temperature is the essential condition for higher production
$NO$.
Thus, option (A) is correct.
Note: In addition temperature, concentration and pressure change also affect the overall system If pressure or concentration increases at equilibrium the reaction will shift towards the direction with lower pressure or lower concentration side and vice versa.
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