Consider the following chemical equilibria: H2 (g) CO2 (g) H2O (g) CO (g) FeO (s) H2 (g) Fe (s) H2O (g) whose equilibrium constants are, respectively, K1=0, 62 and K2=0.42 (at 700 ºC). Calculate the equilibrium constant K3 of the system: FeO (s) CO (g) Fe (s) CO2 (g), at the same temperature.
To calculate the equilibrium constant, K3, of the system FeO (s) + CO (g) ⇌ Fe (s) + CO2 (g), we can use the relationship between the equilibrium constants of different chemical equations.
The equation can be rewritten as the sum of the two given equations:
H2 (g) + CO2 (g) ⇌ H2O (g) + CO (g) (1)
FeO (s) + H2 (g) ⇌ Fe (s) + H2O (g) (2)
First, we need to find the reverse of equation (2) by reversing the equation and taking the reciprocal of its equilibrium constant:
Fe (s) + H2O (g) ⇌ FeO (s) + H2 (g) (2') (K2')
Now we can combine equations (1) and (2') to obtain the desired equation (FeO (s) + CO (g) ⇌ Fe (s) + CO2 (g)). To do this, multiply equation (1) by equation (2'):
(H2 (g) + CO2 (g))(Fe (s) + H2O (g)) ⇌ (H2O (g) + CO (g))(FeO (s) + H2 (g))
Expanding and rearranging, we get:
H2 (g)Fe (s) + CO2 (g)Fe (s) + H2O (g)CO (g) ⇌ H2O (g)FeO (s) + H2O (g)H2 (g) + CO (g)Fe (s)
Simplifying, we have:
Fe (s)(H2 (g)CO2 (g) - CO (g)H2O (g)) ⇌ FeO (s)(H2O (g)H2 (g) - CO (g)H2O (g))
Now, we can use the equilibrium constants K1 and K2' to calculate K3:
K3 = (K1 * K2') / (K1' * K2)
Substituting the given equilibrium constants:
K3 = (0.62 * (1/0.42)) / (1 * 0.42)
K3 = 1.48
Therefore, the equilibrium constant K3 of the system FeO (s) + CO (g) ⇌ Fe (s) + CO2 (g) at 700 ºC is approximately 1.48.