The following chemical reaction is endothermic, requiring [tex]$181 \text{ kJ/mol}$[/tex] of heat:

[tex]$N_2(g) + O_2(g) \longleftrightarrow 2 NO(g)$[/tex]

How would you increase the amount of product produced from this reaction?

A. Heat the reaction and increase the pressure of reactants.
B. Cool the reaction and decrease the pressure of reactants.
C. Heat the reaction and decrease the pressure of reactants.
D. Cool the reaction and increase the pressure of reactants.
E. Temperature and pressure have no effect on product amount.



Answer :

To understand how to increase the amount of product produced in the given endothermic reaction:

[tex]\[ N_{2(g)} + O_{2(g)} \longleftrightarrow 2 NO_{(g)} \][/tex]

we need to consider the principles of Le Chatelier's Principle, which states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.

1. Effect of Temperature:
- Since the reaction is endothermic (ΔH = +181 kJ/mol), it requires heat to proceed.
- Adding heat will favor the formation of products as the system will absorb heat to minimize the disturbance. Therefore, increasing the temperature should shift the equilibrium to the right, increasing the production of [tex]\( NO_{(g)} \)[/tex].

2. Effect of Pressure:
- The reaction equation shows that 1 mole of [tex]\( N_2 \)[/tex] and 1 mole of [tex]\( O_2 \)[/tex] react to form 2 moles of [tex]\( NO \)[/tex].
- In terms of moles of gas, we go from 2 moles of reactants to 2 moles of products, so there is no net change in the number of moles of gas during the reaction. Therefore, changes in pressure will have no significant effect on the position of the equilibrium for this particular reaction.

Given this analysis, the correct strategy to increase the amount of [tex]\( NO_{(g)} \)[/tex] would be to increase the temperature.

Hence, the best approach to increase the amount of product (NO) produced from this reaction is:
- Heat the reaction.

Therefore, the correct answer is:
- Heat the reaction and decrease the pressure of reactants.

However, since pressure does not affect the equilibrium in this case (as the number of gas molecules remains the same), the core factor remains heating the reaction to produce more products.