Question

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.

Answer 1

The equilibrium mole fraction of CO is 0.0515

Step-by-step 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