In an experiment, 0.32 mol H2 and 0.32 mol I2 are mixed in a 1.00-L container, a

Question

In an experiment, 0.32 mol H2 and 0.32 mol I2 are mixed in a 1.00-L container, and the reaction forms HI. If Kc = 49. for this reaction, what is the equilibrium concentration of HI?I2(g) + H2(g) 2HI(g)

Explanation / Answer

H2(g) + I2(g)====>2HI(g) Kc=[HI]²/[H2][I2] (2x)²/((0.30-x)(0.30-x))=49 There's no need for a quadratic. We can just take the square root of both sides because the denominator is (0.30-x)(0.30-x) which is (0.30-x)². So we take the square root of both sides to get 2x/(0.30-x)=7 2x=2.1-7x 9x=2.1 x=2.1/9=2.333 The equilibrium concentration of HI is 2x which is 2*2.333=0.4666 M which is 0.47 M to two significant figures.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

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