The equilibrium constant Kc for the reaction H2(g) + I2(g) equilibrium sign 2Hl(

Question

The equilibrium constant Kc for the reaction

H2(g) + I2(g) equilibrium sign 2Hl(g)

is 54.3 at 430 degrees Celsius. At the start of the reaction there are 0.714 mole of H2, 0.984 mole of I2, and 0.886 mole of HI in a 2.60 L reaction chamber. Calculate the concentrations of the gases at equilibrium.


[H2]=
[I2]=
[HI]=

Explanation / Answer

H2(g) + I2(g) 2Hl(g) initial concentrations are [H2]= 0.714, [I2]= 0.984, [HI]= 0.886 moles in 2.6 L => molarity of [H2]= 0.714/2.6 = 0.2746, [I2]= 0.984/2.6 = 0.3784, [HI]= 0.886/2.6 = 0.3407 now concentrations at equilibrium are [H2]= 0.2746- C , [I2]= 0.3784-C [HI]= 0.3407+ 2C Kc = 54.3 = (2* [HI])^2/ [H2]*[I2] = 4* (0.3407+2C)^2 / (0.2746-C) (0.3784-C) solve for C and substitute in concentration values and u will get the answer

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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