8. Consider the apparatus shown below, in which two compartments are initially s

Question

8. Consider the apparatus shown below, in which two compartments are initially separated by a removable partition (shown by the dashed line). One compartment is filled with nitrogen monoxide and the other with oxygen, with the conditions as shown in the diagram. NO(g) 1.00 L, 600 torr, 127 O2(g) .00L, 500 torr, 127 The partition dividing the container is removed so that the gases mix, the entire container is heated, and the two gases react to produce the maximum possible amount of nitrogen dioxide (NO,, which will also be a gas under these conditions) If the container is at a final temperature of 500. K when the reaction reaches completion, what is the total pressure? (HINT: Be sure you are accounting for the fact that the gases do not just mix, they react with each other!)

Explanation / Answer

moles of gases in each compartment can be calciulate using gas law equation, PV=nRT, n= PV/RT

R= 0.0821 L.atm/mole,K,

1st compartment containing NO, P= 600Torr= 600/760 atm= 0.7894 atm , V=1L, T= 127+273= 400K

n= 0.7894*1/(0.0821*400)= 0.024 moles, moles of O2 in the second compartment = (500/760)*1/(0.0821*400)= 0.020

the reaction betwen NO and O2 is 2NO+O2------->2NO2

theoretical molar ratio of NO: O2= 2:1 actual ratio of NO: O2= 0.024:0.020 = 0.024/0.020 : 1 = 1.2 :1

excess is Oxygen and limiting is NO. moles of NO2 formed= 0.024, Oxygen reacted =0.024/2=0.012

moles of O2 remaining = moles supplied-moles reacted = 0.020-0.012=0.008

total moels of product gases =0.024+0.008 =0.032

n= 0.032, V= 1+1=2L, T= 500K , P= nRT/V= 0.032*0.0821*500/2 = 0.6568 atm

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