Right out of school you accept a job with the environmentally friendly company C

Question

Right out of school you accept a job with the environmentally friendly company Clean Air, Inc. designing diesel engines. They want you to design a hotter engine since your boss thinks that increasing the combustion temperature in the engine from 1000K to 1500K will decrease the amount of NO formed (NO is harmful environmentally since it is metastable in the atmosphere and can be transformed to NO_2 and other toxic compounds). You go to the literature and find out that the combustion reaction is N_2 + O_2 rightarrow 2NO where both reactants and products are gases. Delta G degree for NO at 1000K is 77.7 kJ/mol and is 71.4 kJ/mol at 1500K. The partial pressure of nitrogen gas is 0.69 and that of molecular oxygen is 0.06 in the combustion gas (there is also 17 % CO and 8 % H_2O in the combustion gas which do not enter into the calculation). Calculate the equilibrium concentration of NO at both temperatures. Is your boss right?

Explanation / Answer

N2 + O2 -------- >2NO

first we have to find out equilibrium constant for this reaction

G° = -RTlnK

where G° is standard free energy , given as 77.7KJ /mol at 1000K and 71.4 KJ/mol at 1500K

R is gas constant= 8.314J/(K mol)

T = Kelvin temperature

Rearranging the equation

lnK = -G°/RT

logK = -G°/2.303RT

substituting the value , at 1000K

logK = -(77700(J/mole))/2.303×8.314(J/Kmol)1000K

= - 4.058

K =8.74 × 10^-5

at 1500K the equilibrium constant calculated similarly

logK = - (71400(J/mol)/2.303×8.314( J/Kmol) × 1000KK

= - 2.49

K = 3.23×10^-3

Comparing the K value at 1000 K and 1500 K , at 1500 the reaction will move forward direction and More NO will form

Now we calculate the equilibrium concentration

K = [NO]^2/[N2][O2]

at equilibrium

[N2] = 0.69 - x

[O2] = 0.06-x

[NO] = 2x

Therefore ,at 1000K

8.74 × 10^-5 = (2x)^2/(0.69 -x)(0.06-×)

if we assume 0.69-x = 0.69 and 0.06-×= 0.06 for calculation simplicity

x = 9.51×10^-7

Therefore, partial pressure of NO = 2x = 2×9.51×10^-7 = 1.90×10^-6 at 1000K

at 1500K

3.23 ×10^-3 = (2x)^2/0.69 ×0.06

× = 5.8 × 10^-3

Therefore, partial pressure ofNO at 1500K = 2×5.8×10^-3 = 1.16×10^-2

Comparing the partial pressure value of NO at 1000K and 1500K , we know that at 1500 K partial pressure of NO is increased i.e more NO is forming at 1500 K.So, the boss is not right in his assumption.

  

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