Answered

From beginner to expert, IDNLearn.com has answers for everyone. Ask any question and get a thorough, accurate answer from our community of experienced professionals.

What would happen if [tex]N_2[/tex] were added to [tex]N_2(g) + O_2(g) \rightleftharpoons 2 NO(g)[/tex] at equilibrium?

A. More [tex]O_2[/tex] and NO would form.
B. More NO would form.
C. More [tex]N_2[/tex] and [tex]O_2[/tex] would form.
D. [tex]K_{\text{eq}}[/tex] would decrease.

Sagot :

GinnyAnsweridn
Certainly! Let's analyze the problem step-by-step.

We are given the equilibrium equation:
[tex]\[ N_2(g) + O_2(g) \rightleftharpoons 2 NO(g) \][/tex]

### Understanding Equilibrium

At equilibrium, the rate of the forward reaction (forming NO) is equal to the rate of the reverse reaction (forming [tex]\(N_2\)[/tex] and [tex]\(O_2\)[/tex]). According to Le Chatelier's principle, if a system at equilibrium is disturbed, the system will adjust itself to counteract that disturbance and restore a new equilibrium.

### Applying Le Chatelier's Principle

If additional [tex]\(N_2\)[/tex] is added to the system, the concentration of [tex]\(N_2\)[/tex] increases. Le Chatelier's principle tells us that the system will respond by shifting the equilibrium to the right to reduce the effect of the added [tex]\(N_2\)[/tex]. This shift to the right means that more [tex]\(NO\)[/tex] will be produced as [tex]\(N_2\)[/tex] and [tex]\(O_2\)[/tex] react.

### Analyzing the Options

Let’s evaluate each option:

A. More [tex]\(O_2\)[/tex] and [tex]\(NO\)[/tex] would form.
- This is incorrect because adding [tex]\(N_2\)[/tex] will not independently increase the concentration of [tex]\(O_2\)[/tex], but will instead drive the reaction to form [tex]\(NO\)[/tex].

B. More [tex]\(NO\)[/tex] would form.
- This is correct. Adding [tex]\(N_2\)[/tex] shifts the equilibrium to the right, resulting in the formation of more [tex]\(NO\)[/tex].

C. More [tex]\(N_2\)[/tex] and [tex]\(O_2\)[/tex] would form.
- This is incorrect because adding [tex]\(N_2\)[/tex] will not shift the reaction to the left to form more [tex]\(N_2\)[/tex] and [tex]\(O_2\)[/tex]. Instead, it shifts to the right, forming more [tex]\(NO\)[/tex].

D. [tex]\(K_{\text{eq}}\)[/tex] would decrease.
- This is incorrect. [tex]\(K_{\text{eq}}\)[/tex], the equilibrium constant, is dependent only on temperature and is not affected by changes in the concentrations of reactants or products.

### Conclusion

Based on the equilibrium principles and the shift in the reaction due to the addition of [tex]\(N_2\)[/tex], the correct answer is:

B. More [tex]\(NO\)[/tex] would form.
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your search for answers ends at IDNLearn.com. Thank you for visiting, and we hope to assist you again soon.