CO (g) H20 (g) H2 (g) +CO2 (g) 1- Using information in the Thermodynamic Data ta
ID: 1060242 • Letter: C
Question
CO (g) H20 (g) H2 (g) +CO2 (g) 1- Using information in the Thermodynamic Data table on D2L, calculate the equilibrium constant K for that reaction at 298 K. 2- Using information in the Thermodynamic Data table on D2L, calculate AH for that reaction 3- Calculate the equilibrium constant K for that reaction at 1150 K assuming AH is independent temperature 4- 1 mole of CO (g) and 1 mole of H20 (g) are introduced in a reaction vessel at 1150 Kand establish equilibrium. Calculate the number of moles of each species at equilibrium. 5- The amount of CO2 in the equilibrium mixture is doubled and the system is allowed to reach a new equilibrium at 1150 K. Calculate the number of moles of each species in the new equilibrium conditions.Explanation / Answer
The given reaction is
CO (g) + H2O (g) <=====> H2 (g) + CO2 (g)
1) Look up the table and compute the free energy change of the reaction, G0r as
G0r = n.G0f (products) – n.G0f (reactants)
Pick up values of G0f from the table and write:
G0r = [(1 mol H2)*(0 kJ/mol) + (1 mol)*(-394.359 kJ/mol)] –[(1 mol)*(-137.168 kJ/mol) + (1 mol)*(-228.59 kJ/mol)] = (-394.359 kJ) – (-137.168 kJ – 228.59 kJ) = (-394.359 kJ) + 365.758 kJ = -28.601 kJ
(I had to pick up few values from the internet since your supplied table doesn’t contain these values).
We know that
G0r = -R*T*ln K where K is the equilibrium constant. Plug in T = 298 K and the value of G0r from above to obtain
(-28.601 kJ) = -(8.314 J/mol.K)*(298 K)*ln K
====> -28.601 kJ = (-2.477 kJ)*ln K
====> ln K = 11.5466 11.547
====> K = e^(11.547) = 103466.17 1.035*105 (ans).
2) Look up the table (also few values from internet sources) and compute H0r as
H0r = n.H0f (products) – n.H0f (reactants) = [(1 mol)*(0 kJ/mol) + (1 mol)*(-393.509 kJ/mol)] – [(1 mol)*(-110.525 kJ/mol) + (1 mol)*(-241.82 kJ/mol)] = (-393.506 kJ) – (-110.525 kJ – 241.82 kJ) = -393.509 kJ + 352.345 kJ = -41.164 kJ (ans).
3) Use Vant Hoff equation to obtain
ln (K2/K1) = -H0r/R*(1/T2 – 1/T1) where K1 = equilibrium constant at T1 = 298 K, K2 = equilibrium constant at T2 = 1150 K and H0r = -41.164 kJ. Plug in values to obtain
ln (K2/1.035*105) = -(-41.164 kJ/8.314 J/mol.K)*(1/1150 – 1/298) K-1
===> ln (K2/1.035*105) = (41.164*103/8.314)*(-2.486*10-3)
===> ln (K2/1.035*105) = -12.3086
===> (K2/1.035*105) = e^(-12.3086) = 4.51276*10-6
===> K2 = 1.035*105*4.51276*10-6 = 0.467 0.47 (ans).
4) Set up the ICE chart for the reaction at equilibrium:
CO (g) + H2O (g) <====> H2 (g) + CO2 (g)
Initial 1 1 0 0
Change -x -x +x +x
Equilibrium (1 – x) (1 – x) x x
K = [CO2][H2]/[CO][H2O] = (x).(x)/(1-x)(1-x)
At 1150 K, we have
0.47 = x2/(1 – x)2
====> 0.47 = x2/(1 – x)2
Take positive square root on both sides and write
0.68556 = x/(1 - x)
===> 0.68556 – 0.68556x = x
===> 0.68556 = 1.68556x
====> x = 0.4067 0.41
Therefore, [CO] = (1.0 – 0.41) = 0.59 mol.
[H2O] = (1.0 – 0.41) = 0.59 mol.
[H2] = 0.41 mol.
[CO2] = 0.41 mol (ans).