OWLv2 | Online teaching and learming resource from Cengage Learning Use the Refe

Question

OWLv2 | Online teaching and learming resource from Cengage Learning Use the References to access impertant values if needed far this gurstion. A student ran the following reaction in the laboratory at 722 K Hzfg) +12(g) ? 2111E) When she introduced 0.212 moles of HE) and 0.215 moles of I(e) into a 1.00 lter container, she found the equilibrium concentration of Hl) to be 0.348 M Calculate the equilibrium constant, K, she obtained for this reaction Retry Entire Group 4 more group attempts remaining

Explanation / Answer

The reaction with ICE TABLE

H2 + I2 = 2HI

I 0.212 0.235

C (-x) (-x) +2x

E (0.212-x) (0.235-x) 2x

At equilibrium

[HI] = 0.348 = 2x

x = 0.174

[H2] = 0.212-x = 0.212 - 0.174 = 0.038 M

[I2] = 0.235-x = 0.235 - 0.174 = 0.061 M

Equilibrium constant expression of the reaction

Kc = [HI]2 / [H2] [I2]

= 0.348*0.348 / (0.038*0.061)

= 52.245

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

---

---

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