1.Moles of NO2 initially= mass/molar mass , molar mass of NO2= 46 Moles of NO2 i

Question

1.Moles of NO2 initially= mass/molar mass , molar mass of NO2= 46

Moles of NO2 initially = 7.18/46=0.16 ,

Let x= moles of N2O4 formed to reach a pressure of 0.75 atm

Moles : NO2= 0.16-2x and N2O4= x, total moles = 0.16-2x+x=0.16-x

From gas law equation, moles at that time, n= PV/RT, P=0.75 atm, V=3.5L, T= 300K and R=0.0821 L.atm/mole.K, n= 0.75*3.5/(0.0821*300)=0.11

Hence 0.16-x=0.11 or x=0.16-0.11=0.05, total moles corresponding to that pressure =0.16-0.05=0.11

Moles : NO2=0.16-2*0.05=0.06 and N2O4=0.05

Mole fraction = moles/total moles, mole fractions: NO2=0.06/0.11=0.545 and N2O4=1-0.545=0.455

2.

Moles= mass/molar mass, molar masses : Cl2= 70.90 and ethylene(C2H4)= 28.05

Moles : C2H4= 11.22/28.05=0.400 and Cl2= 35.45/70.9=0.5

The reaction is C2H4+Cl2---->C2H4Cl2, Molar ratio of C2H4:Cl2= 1:1

Actual molar ratio of C2H4:Cl2= 0.4:0.5, dividing by 0.4, the lowest value, the ratio becomes

C2H4:Cl2= 0.4/0.4:0.5/0.4 =1:1.25

Excess reactant is Cl2. When percent yield is 100%, all the limiting reactant C2H4 reacts and as per the reaction, 0.4 moles of C2H4Cl2 forms. Molar mass of C2H4Cl2=70.9+28.05=98.95 g/mole

Mass of C2H4Cl2 formed for 100% yield = 0.4*98.95=39.58 gm

When the yield is only 50%, moles of C2H4 reacted=0.4*0.5=0.2, moles of Cl2 reacted =0.2 ( since one mole of limiting reactants C2H4 requires 1 mole of Cl2 for forming C2H4Cl2)

Moles of C2H4 remaining = 0.4-0.2=0.2 moles and moles of Cl2 remaining =0.5-0.2=0.3

Masses of reactants (gm): C2H4= 0.2*28.05=5.61 and Cl2= 0.3*70.9=21.27

Total mass of reactants after the reaction = 5.61+21.27=26.88 gm

3. work done = -integral of PdV= -deltan*R*T

deltan= change in no of moles during the reaction.

moles of Reactants : H2= 3.5 and O2= 1.5, moles of H2O formed based on O2, the limiting reactant = 2*1.5= 3moles

moles of H2 reacted= 2*1.5=3, moles of H2 remaining= 0.5

so total moles of products, deltan= 3.5

change in moles during the reaction = 3.5-(3.5+1.5)=-1.5

work done = 1.5*8.314*298=3716 joules

deltaH= deltaU+ PdV+V*deltaP

at constant pressure, deltaP=0

deltaH = -483.6 Kj/mole= -483.6*1000J/mole

-483.6*1000= deltaU+3716

deltaU= -483.6*1000-3716=-487316 Joules=-487.316 KJ

Explanation / Answer

(VIII) (10 points) Nitrogen dioxide dimerizes to give dinitrogen tetroxide Page 7 of 8 2 NO2(g) N2049 At 300 K, initially, 7.18 g of NO, (no N 04) is added into a 3.50-L closed container. After a certain period, the pressure of the closed container is measured to be 0.75 atm. What are the mole fractions of the two gases NO /N20s mixture? (Hint: Initially, the system contains only NO2. After a certain per mixture of NO /N:04.) in the riod, the system contains a (1X) (10 points) The dry-cleaning solvent 1,.2 dichloroethane (also called ethylene chloride), C-HaCh, is prepared by reaction of 11.22 g of ethylene (C2H4) with 35.45 g of Cl2. (a) If the percent yield is 100.0%, after the reaction, how many grams of the product of 1,2 dich! produced? (b) If the percent yield is 50.00%, after the reaction, what is the total mass of the reactants left over? (Molar mass of C2H4-28.05 g mol; Molar mass of CI, = 70.90 g mol; Molar mass of Clich-98.95 gmol oroethane Page 6 of S II) (10 points) Consider the reaction, 2H2(g) + O2(g) 2H20(g) and 1.50 moles of Horn=-483.6 k7mol 3.50 moles of H2 (g) O:(g) are converted to H O(R) against a pressure of 1.00 atm at 25.0°c. :0(g) (a) Calculate the work (w) done by this reaction. (b) what is the change reaction? Assume that the reaction goes to completion. in internal energy (AU) for this

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