The reaction for synthesis of ammonia is CO + 3H 2 CH 4 + H 2 O At 10 atm and 40

Question

The reaction for synthesis of ammonia is CO + 3H2 CH4 + H2O

At 10 atm and 400 oC, it was observed that 3.85% of the gas mixture was NH3, starting initially from pure NH3. What is the value of KP?

The reaction for synthesis of ammonia is

begin inline style 3 over 2 end style H subscript 2 left parenthesis g right parenthesis space plus space begin inline style 1 half end style N subscript 2 left parenthesis g right parenthesis space equals N H subscript 3 left parenthesis g right parenthesis

At 10 atm and 400 oC, it was observed that 3.85% of the gas mixture was NH3, starting initially from pure NH3. What is the value of KP.

Hint: Construct the reaction chart.

0.0128

0.1280 is wrong

1.280

Explanation / Answer

the reaction is 2NH3(g)<----->N2(g)+ 3H2(g)

let mole of NH3= 2 ( pure NH3, let x= mole of N2 formed to reach equilibrium

hence at equilibrium NH3= 2-2x, N2=x and H2=3x

total moles at equilibrium = 2-2x+x+3x= 2+2x

mole fractions at equilibrium, NH3= (2-2x)/(2+2x)= (1-x)(1+x)= 3.85/100, N2=x/(2+2x), H2= 3x/(2+2x)

hence 1-x= 0.0385*(1+x)

1-0.0385= x*(1+0.0385)

x= 0.93, hence at equilibrium, mole fraction NH3 =3.85/100 =0.0385, N2= 0.93/(2+2*0.93)=0.241 and H2= 3*0.93/(2+2*0.93)=0.72

partial pressures = mole fraction* total pressure, Partial pressures: NH3= 0.0385*10 =0.385atm, N2= 0.241*10=2.41 atm and H2=0.72*10=7.2 atm

Kp= PH23*PN2/ PNH32, P =partial pressure of component

KP= 7.23* 2.41/(0.3853)= 2342

x=

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