Answer:
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
