The reaction is O2 + 2H2 -> 2H2O Starting with 1 mole of O2 and 4 mol H2, let th

Question

The reaction is O2 + 2H2 -> 2H2O Starting with 1 mole of O2 and 4 mol H2, let the reaction goto completion. The reaction vessel is in 11.2 L and I keeptemp at 25 C. What is the pressure A) before and B) after the reaction runs to completion. Assume ideal (perfect)gas behavior. HINT: The pressure after (pt b) should be 8.74atm Will rate asap! The reaction is O2 + 2H2 -> 2H2O Starting with 1 mole of O2 and 4 mol H2, let the reaction goto completion. The reaction vessel is in 11.2 L and I keeptemp at 25 C. What is the pressure A) before and B) after the reaction runs to completion. Assume ideal (perfect)gas behavior. HINT: The pressure after (pt b) should be 8.74atm Will rate asap!

Explanation / Answer

The reaction is                                                    O2 + 2H2 -> 2H2O no. of moles atstart                    1       4            0 no.of molesreacted                    1        2           2 no. of moles left afterRXN         0     (4-2)         2 Volume of the vessel is,V= 11.2 L temp. T = 25 o C = 25 + 273 = 298 K pressure before reaction ------------------------- From gas law PV = nRT n = No. of moles before the reaction = 1 + 4 = 5 moles V = Volume = 11.2 L R = gas constant = 0.0821 L atm / mol-K T = temp = 298 K Substituting values we get P = nRT / V                                           = (5* 0.0821 * 298 ) / 11.2                                           = 10.922 atm pressure after reaction ------------------------- From gas law PV = nRT n = No. of moles before the reaction = (4-2) + 2 = 4moles V = Volume = 11.2 L R = gas constant = 0.0821 L atm / mol-K T = temp = 298 K Substituting values we get P = nRT / V                                           = (4* 0.0821 * 298 ) / 11.2                                           = 8.7377 atm                                            ~ 8.74 atm ------------------------- From gas law PV = nRT n = No. of moles before the reaction = (4-2) + 2 = 4moles V = Volume = 11.2 L R = gas constant = 0.0821 L atm / mol-K T = temp = 298 K Substituting values we get P = nRT / V                                           = (4* 0.0821 * 298 ) / 11.2                                           = 8.7377 atm                                            ~ 8.74 atm

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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