Question

Problem 14.133 20 of 20 ConstantsI Periodic Table ? Part A At 100° C, Kc 4.72 for the reaction 2NO2(9)N204(9) An empty 10.0 L flask is filled with 1.15 g of NO2 at 100 C What t is the total pressure in the flask at equilibrium Potal-0.0504 atm Submit ious Ans Request Ans X Incorrect Try Again < Return to Assignment Provide Feedback

Answer 1

Kc = 4.72

2NO2(g) ---------------- N2O4(g)

$\Delta$ n = number of moles of gaseous products - number of moles of gaseous reactants

$\Delta$ n = 1-2 = -2

T= 100C = 100+273= 373K

R= 0.0821 L-atm/moleK

Kp= Kc (RT)^ $\Delta$ n

Kp= 4.72x(0.0821x373)^-1

Kp= 0.154

mass of NO2 = 1.15gram

molar mass of NO2= 46.0gram/mole

number of moles ofNO2 = 1.15/46.0= 0.025 moles

T= 100C=373K

Volume= V= 10.0L

R= 0.0821L-atm

PV=nRT

P=nRT/V= 0.025x0.0821x373/10.0= 0.0766

Partial pressure of NO2= 0.0766 atm

2 NO2 --------------- N2O4

0.0766                    0

-2x                         +x

0.0766-2x              +x

Kp = PN2O4/P^2NO2

0.154 = x/(0.0766-2x)^2

for solving the equation

x=0.00086

Pressure of NO2 = 0.0766-2x= 0.0766-2(0.00086) = 0.07488 atm

pressure of N2O4 = x= 0.00086 atm

Total pressure = 0.07488+0.00086= 0.07574 atm

Total pressure = 0.0757 atm