Consider the apparatus shown below in which two 100L bulbs a

Consider the apparatus shown below, in which two 1.00-L bulbs are initially separated from one another by a closed valve. One bulb is filled with 494 torr of sulfur dioxide and the other with 396 torr of oxygen, and the entire apparatus is at an initial temperature of 23 C Before mixing 02 1.00 L SO2 V = 1.00 L P= 4.94 x 102 torr P= 3.96 × 102 torr T= 23°C T = 23° C The valve separating the bulbs is opened so that the gases mix, the entire container is heated, and the two gases react to produce the maximum possible amount of SOs. If the container is at a final temperature of 561 K when the reaction reaches completion, what is the total pressure? Pressure torr

Solution

molres of each gas can be calculated from gas law equation,

PV= nrT, P= pressure ain atm , V= volume in L, T= temperature in K, n= no of moles and R= 0.0821L.atm/mole.K

since 760 Torr= 1 atm

for SO2, P= 4.94*100/760 atm =0.65 atm, V= 1L, T= 23deg.c= 23+273= 296K, n= 0.65*1/(0.0821*296)= 0.027

for O2 , P= 3.96*100/760 atm =0.52 atm, V= 1L, T= 23 deg.c= 296K, n= 0.52*1/(0.0821*296)=0.0214

the reaction between SO2 and O2 is SO2+0.5O2------>SO3, theoretical molar rati of SO2: O2= 1:0.5

actual molar ratio of SO2:O2= 0.027:0.0214 =0.027/0.0214: 1= 1.26:1 = 1.26/2 :1/2= 0.63:0.5

so limiting reactant is SO2 and all the SO2 is consumed. moles of SO3 formed =0.027 ( from the reaction, 1 mole of SO2 gives 1 moles of SO3). moles of oxygen consumed= 0.027/2=0.0135, moles of O2 reamining =0.0214-0.0135 =0.0079

moles of SO3 and O2= 0.027+0.0079 =0.0349, this is final n= 0.0349, V= total volume after reaction = 1+1= 2L

T= 561K, P= nRT/V= 0.0349*0.0821*541/2=0.775 atm=0.775*760 torr=589 Torr

P=