The combustion products from burning pentane. CSH I2, with pure oxygen in a stone- .stoichiometric ratio exit at 2400 K, 100 kPa. Consider the dissociation of only CO2 and find the equilibrium mole fraction of CO.

The equilibrium mole fraction of CO is 0.0515

Explanation:

Solution

Given that:

Temperature T = 2400 K

Pressure P = 100 kPa

The reaction C₅H₁₂ + 8 O₂ →5CO₂ + 6 H₂ O

Now,

We take in K value from logarithm to the base e of the equilibrium constant K at 2400 K

From reaction 2CO₂⇔ 2 CO + O₂

In K = -7.715

P° =0.1 MPa

K = e^⁻7.715

=4.461 * 10^⁻4

Thus,

2CO₂⇔ 2 CO + IO₂

The initial mole CO₂ and shift reaction with 'x'

The  atom balance from the above reaction is given below:

Species: CO₂ CO O₂ H₂O

Initial:  5006

Changes: - 2x 2xx0

Total:5 - 2x 2xx6

The total number of moles n tot = 5-2x+2x+x+6

= 5+x+6

=11+x

Now,

from the equation of the equilibrium constant K

K =Y²co₂Yo₂/Y²co₂ (P/P₀)

Yc₀ =nc₀/n tot

= 2x/11+x

Yco₂ = nc₀/n tot

= 5-2x/11 +x

So,

K =(2x/1+x)² *x/1+x/(5-2x/11 +x)² * (100/100)

4.461 * 10^⁻4 = (2x)²/(11 +x)² x/(11+x) (11+x)/(5-2x) *1

4.461 * 10^⁻4= (2x/5-2x)² x/11+x

Now by trial and error method by keeping  x =0.291 the value satisfies the equation

4.461 * 10^⁻4 =(2 *0.291/5-(2* 0.291))² *(0.291/11 + 0.291)

4.461 * 10^⁻4 =4.47 *10^⁻4

Hence

x =0.291

Thus,

The mole fraction of CO₂ =nco₂/ntot

Yco₂ = 5-2x-/11 +x

Yco₂ = 5-(2 * 0.291)/11+0.291

Yco₂ = 0.39128

So,

Mole fraction of CO =nco/ntot

Yco = 2x/11+x

=2 * 0.291/11+0.291

Yco =0.0515

Then,

Mole fraction of O₂ =no₂/ntot

Yo₂ = x/11 +x

= 0.291/11+0.291

=0.02577

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