Question

According to Le-Chatelier’s principle, an increase in the temperature of the following reaction will
${{N}_{2}}+{{O}_{2}}\rightleftharpoons 2NO-43,200$kcal [MP PMT $1985,93$]
A.Increase the yield of $NO$
B.Decrease the yield of $NO$
C.Not affect the yield of $NO$
D.Not help the reaction to proceed in the forward direction

Answer 1

Hint: According to Le-Chatelier’s principle, when temperature increases the system will try to compensate for that change and reestablish the equilibrium. The given reaction is endothermic and will be favored by an increase in temperature in the forward direction.

Complete step-by-step solution:When a chemical reaction is at equilibrium state and reaction conditions are changed, the reaction system is no longer in an equilibrium state. Then the system will try to reach a new equilibrium that neutralizes the change in conditions.
Le-Chatelier’s principle describes what happens to a system when something changes and takes it away from equilibrium. There are three ways in which we can change the conditions of a reaction system at equilibrium:
a.Changing the temperature of the system
b.Changing the concentrations of the constituents present in the reaction
c.Changing the pressure of the reaction system
Changes in the temperature of the system affect the equilibrium state by changing the value of the equilibrium constant for the reaction. Increasing the temperature in an endothermic reaction, equilibrium will shift in that direction where it can reduce the increased temperature.
Here the given reaction is:${{N}_{2}}+{{O}_{2}}\rightleftharpoons 2NO-43,200kcal$
Or, ${{N}_{2}}+{{O}_{2}}+43,200\,\,kcal\rightleftharpoons 2NO$
It is an endothermic reaction in which reactants absorb $43,200kcal$heat to form nitrous oxide, $NO$. With the increase in temperature, the reaction equilibrium will shift in the forward direction to maintain the overall temperature of the system. So, more ${{N}_{2}}\,\And {{O}_{2}}$ reacts to reduce the increased temperature of the system, as a result, the yield of the product, $NO$ increases.

Thus, option (A) is correct.

Note: The equilibrium constant varies with temperature. The magnitude of the equilibrium constant of an endothermic reaction increases while for an exothermic reaction the value of the equilibrium constant decreases.