If we represent the equilibrium as: 2 NO_2(g) N_2O_4(g) We can conclude that: Th

Question

If we represent the equilibrium as: 2 NO_2(g) N_2O_4(g) We can conclude that: This reaction is: A. Exothermic B. Endothermic C. Neutral D. More information is needed to answer this question. When the temperature is decreased the equilibrium constant. K: A. Increases B. Decreases C. Remains the same D. More information is needed lo answer this question. When the temperature is decreased the equilibrium concentration of N_2O_4: A. Increases B. Decreases C. Remains the same D. More information is needed to answer this question.

Explanation / Answer

Ans. 1. Correct option. C. Neutral

NO2 is brown in color and N2O4 is colorless. As shown in figure, the brown color disappears on cooling. It means brown-colored N2O is converted into colorless N2O4. SO, the reaction goes in forward direction upon cooling.

Thus, NO2 dimerizes into N2O4 when heat is lost (cooling = loss of heat). Or, N2O4 dissociates into N2O by absorbing heat (endothermic reaction).

Since the reaction is at equilibrium, no net heat is gained or lost. So, the reaction is neutral (no heat exchange) at equilibrium.

Ans. 2. Le Chatelier’s principle “if a dynamic equilibrium is disturbed by changing the conditions (Concentration, Volume, Pressure, temperature, etc.), the position of equilibrium shifts to counteract the change to reestablish an equilibrium”

When temperature is decrease, more N2O4 forms.

So, equilibrium constant, K = [N2O4] / [N2O]2 becomes greater than (increases) initial value because [N2O4] has increased.

Correct option. A. Increase

Ans. 3. Correct option. A. Increase             [see Ans.2]

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.