Determine the Kc value for the following reaction if the equilibrium concentrati

Question

Determine the Kc value for the following reaction if the equilibrium concentrations are : [HBr] = 0. 20 M [H2] = 0.12 M 2HBr (g) = H2 (g) + Br2 (g) 6.0 Times 10^-3 0.30 16.7 3.33 7.2 Times 10^-4 Consider the following reaction, equilibrium concentrations. and equilibrium constant at a particular temperature. Determine the equilibrium concentration of H2O(g). C2H4(g) + H20(g) = C2H5OH(g) Kc = 9.0 Times 10^3 [C2H4)eq = 0.015 M [C2H5OH)eq = 1.69M 9.9 Times 10^-7 M 80.M 1.0 M 1.68M 0.013 M Consider the following reaction: Xe(g) + 2 F2(g) rightarrow XeF4(g) A reaction mixture initially contains 2.24 M Xc and 4.27 M F2-concentration of Xc is 0.34 M, find the equilibrium constant (Kc) for rthe reaction. 0.12 0.99 25 8.3 0.040

Explanation / Answer

ANSWER:

Q19. Kc = [H2][Br2]/[HBr]2

Kc = 0.1 X 0.12/ [0.2]2 = 0.3

Q20.

Kc = [C2H5OH]/[C2H4][H2O]

9.0 X 103 = 1.69/0.015[H2O]

[H2O] = 1.69/0.015 X 9.0 X 103 = 0.0125 = 0.013M ( after rounding off)

Q21.

Xe + 2F2 ,<--------------> XeF4

1 mole of Xe reacts with 2 moles of F2.

Amount of xe rected = initial concentration - equilibrium concentration = 2.24 - 0.34 = 1.9M

moles of F2 reacted = 2 X 1.9 = 3.8

equilibrium concentration of F2 = initial concentration - moles of F2 reacted = 4.27 - 3.8 = 0.47

One mole of Xe gives one mole of XeF4 . Therefore 1.86 moles of Xe will give 1.9 moles of XeF4

Kc = [XeF4]/[Xe][F2]2

Kc = 1.9/0.34 X (0.47)2 = 25.2 = 25

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

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