When the reversible reaction N_2 (g) + O_2 (g) 2 NO_(g) has reached a state of e

Question

When the reversible reaction N_2 (g) + O_2 (g) 2 NO_(g) has reached a state of equilibrium (A) no further reaction occurs (B) the total moles of products must system the remaining moles of reactant. (C) the concentration of each substance in the system will be constant. (D) the addition of a catalyst will cause formation of more NO. tetrafluoride, XeF_4 can be prepared by heating Xe and F_2 together as represented by Xe(g) + 2 F_2 (g) XeF_4 (g) What is the equilibrium constant expression for this reaction? (A) K_c = [XeF_4]/[Xe] [F_2] (B) K_c = [Xe] [F_2]/[XeF_4] (C) K_c = [XeF_4]/[Xe] [F_2]^2 (D) K_c = [XeF_4]/2[Xe] [F_2] numerical value of the equilibrium constant for any chemical change is by changing the A) nature of the catalyst. (B) concentration of the products. (C) the pressure. (D) the temperature.

Explanation / Answer

1ans:C, the state of the reversible reaction at which the concentrations of the reactants and products do not change with time is called the equilibrium state.

2.ans: C: Kc= [XeF4]/ [Xe][F2]2

3.

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

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