at 298K, S2F4(g) partially decomposes to SF2 (g). atequilibirum the partial pres

Question

at 298K, S2F4(g) partially decomposes to SF2 (g). atequilibirum the partial pressure of SF2(g) is 8.4x10-2torr and the partial pressure of S2F4(g) is 36.8 torr. S2F4(g)<-->2SF2(g) determine the thermodynamic equilibirum constant for thisdecomposition at 298K, S2F4(g) partially decomposes to SF2 (g). atequilibirum the partial pressure of SF2(g) is 8.4x10-2torr and the partial pressure of S2F4(g) is 36.8 torr. S2F4(g)<-->2SF2(g) determine the thermodynamic equilibirum constant for thisdecomposition

Explanation / Answer

S2F4(g) <-->   2 SF2(g) Kp = [SF 2] 2 / [ S2 F4]       = (8.4 * 10-2)2 / (36.8)       = 1.91 *10-4 Change in number of gaseous moles, n = 2- 1 = 1 Kc = Kp (RT) n       = 1.91 * 10-4 *( 0.0821 L-atm/mol/K * 298 K) 1       = 0.00469       = 0.00469

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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