Long Answer PROBLE 2N0(g) + O2 (g) react. Determine the partial pressures of gas

Question

Long Answer PROBLE 2N0(g) + O2 (g) react. Determine the partial pressures of gases remaining at the end of the reaction and heat is absorbed or released during the reaction based on the reagents given? M 110 marks): Nitric oxide (NO) reacts with molecular O2 as follows: 2NO2 (g) . When the valve is opened, the gases are then allowed to mix and No O 4.00 L at 0.500 atm 2.00 L at 1.00 atm Here is the data you'll need: > Temperature remains constant at 25 oC > Hof values can be found in your text/workbook

Explanation / Answer

PV = nRT

n = PV/(RT) = 0.5*4/(0.082*298) = 0.0818 mol of NO

n = PV/(RT) = 2*1/(0.082*298) = 0.0818 mol of O2

raito is 2:1 so there is ecxes NO

mol of O2 left --> 0

mol of NO2 formed = 0.0818 *0.5 =  0.0409 mol of NO2 formed

mol of NO left = 0.0409

total mol = NO + O2 + NO2 = 0.0409 + 0 + 0.0409 = 0.0818

PV = nRt

P = nRT/V = (0.0818 *0.082*298)/(4+2)

P = 0.3331 atm

P-NO = 0.5*0.3331 = 0.16655 atm

P-NO2 =0.5*0.3331 = 0.16655 atm

HRXN = hproduct s- hreactants

HRXN = 2NO2 - (2NO + O2)

HRXN = 2*33.2 - (2*91.3 +0) = -116.2 kJ per mol of O2

mol of O2 reacted = 0.0818

Q = 0.0818*-116.2

Q = -9.50516 kJ

Q = 9.50516 KJ released

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

---

---

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