A gaseous species A decomposes into gaseous species B and C as follows 2 A_gas r

Question

A gaseous species A decomposes into gaseous species B and C as follows 2 A_gas rightarrow 2B_gas + C_gas The dissociated proportion of A increases with the temperature. 0.4 mol of A at 20 degree C is put in a reactor of 61 that was emptied down to vacuum before. Then, it is heated. At 482 degree F, the pressure in the reactor is 3.3 atm. Show that this observation proves the existence of a chemical reaction. What is, at 482 F, the proportion (in %) of A that has been dissociated? What are the partial pressures of the different gases in the mixture at 482 degree F? At 1472 degree F, the dissociation of A is total. Calculate the pressure in the reactor. Which would be the pressure at 1000 degree C?

Explanation / Answer

1) consider the equation of ideal gas ,PV=nRT ,P=pressure of gas,V=volume =6 L,n=moles of gas=0.4moles

R=universal gas constant=0.0821L atm/K.mol

Ti=temperature=20+273=293K

Tf=482 deg F=(482-32)*5/9=250 deg C=250+273=523K

we can estimate the pressure at 293K

P=nRT/V=0.4 mol*0.0821Latm/K.mol *293K/6L=1.6 atm

As pressure of gas increased at Tf so there must be chemical rxn going on increasing gases in the reactor ,thereby increasing the pressure.

2) calculate the new n''

PV=n"RT

n"=PV/RT=3.3atm *6L/0.0821 Latm/K.mol *523K=0.46 moles

So excess moles =0.46-0.4=0.06moles =moles of gas formed

As 2 moles of A produces 3 moles of (B&C) ,so 1 mol A dissociates to produce 3/2=1.5 moles

or,1.5 moles of B&C is produced by 1 mol A dissociation

or,0.06 moles of B&C is produced by 0.06/1.5 =0.04 moles of A

thus,0.04 moles of A has dissociated.

fraction of A dissociated=0.04/0.4 =0.1

% A dissociated=0.1*100=10%

3)0.04 moles of A forms 0.04 moles of C and 0.04*2=0.08 moles of B

mole fraction of B=0.08/total moles=0.08/0.46=0.2

mole fraction of C=0.04/0.046=0.09

So partial pressure of A=mole fraction*total pressure (Raoult's law)=0.2*3.3atm=0.66 atm

pB=0.09*3.3atm=0.033 atm

4)Tf=1472 def F=(1472-32)*5/9=800 deg C=800+273=1073K

if 0.4 moles of A dissociates,moles of C formed=0.4 moles

and moles of B formed=0.4*2=0.8moles

total moles of gas formed=0.4+0.8=1.2 moles

P=nRT/V=1.2moles*0.0821Latm/K.mol*1073K/6L=17.6 atm

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