The equilibrium constant, K, for the reaction H_2(g) + F_2(g) 2HF(g) has the val

Question

The equilibrium constant, K, for the reaction H_2(g) + F_2(g) 2HF(g) has the value 2.1 X 103 at a particular temperature. When the system is analyzed at equilibrium at this temperature, the concentrations of H2(g) and F2(g) are both found to be 0.0021 M. What is the concentration of HF(g) in the equilibrium system these conditions? The reaction H2(g) + I_2(g) 2HI(g) has K_p = 45.9 at 763 K. A particular equilibrium mixture at that temperature contains gaseous HI at a partial pressure of 4.00 atm and hydrogen gas at a partial pressure of 0.200 atm. What is the partial pressure of I2?

Explanation / Answer

Kc = [HF]^2/[H2][F2]

2.1*10^3 = (2X)^2/(0.0021-X)^2

X = 0.002 M

at equilibrium

Concentration OF HF = 0.002*2 = 0.004 M

Concentration OF H2 = 0.0021-0.002 = 0.0001 M

Concentration OF F2 = 0.0021-0.002 = 0.0001 M

42)

Kp = pHI^2/pH2*pI2

45.9 = (4^2/(0.2*x))

x = 1.743

partial prerssure of I2 = 1.743 atm

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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