Ideal Gas Law and Applications A flask of volume 5.00 L is evacuated and 43.78 g

Question

Ideal Gas Law and Applications A flask of volume 5.00 L is evacuated and 43.78 g of solid dinitrogen tetroxide, N_2O_4, is introduced at -19 degree C. The sample is then warmed to 25 degree C, during which time the N_2O_4 vaporizes and some of it dissociates to form brown gas NO_2. The pressure slowly increases until it stabilizes at 2.96 atm. Please write a balanced equation for this reaction. If the gas in the flask at 25 degree C were all N­_2O_4, what would the pressure be? If all the gas in the flask were converted into NO_2, what would the pressure be? What are the mole fractions of N_2O_4 and NO_2 once the pressure stabilizes at 2.96 atm?

Explanation / Answer

the reaction is N2O4--->2NO2

b)molar mass of N2O4, n = 28+64= 92 , moles of N2O4 in 43.78 gm = 43.78/92=0.48 moles, T= -19+273= 254K

V= 5L, R= 0.0821 L.atm/mole.K, Pressure = nRT/V= 0.48*0.0821L.atm/mole.K* 254/5=2atm

c) moles of NO2 formed when all the N2O4 got converted = 0.48*2= 0.96 ( 1 mole of No2 gives 2 mole of NO2)

then the pressure = 2* pressure only when N2O4 is there = 2*2=4atm

d) let x= moles of N2O4 reacted, moles of NO2 formed= 2x, moles of NO2 remaining = 0.48-x

total moles = 0.48-x+2x= 0.48+x

P= 2.96= (0.48+x)*0.0821*254/5

0.48+x= 2.96*5(/0.0821*254)=0.71

x= 0.71-0.48=0.23

moles of N2O4= 0.48-0.23= 0.25 moles of NO2=2*0.23=0.46

total moles = 0.25+0.26=0.51, mole fraction : moles/total moles : N2O4= 0.25/0.51= 0.49 and NO2= 1-0.49=0.51

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