What is K of Fe(S) + H2O(g)=FeO(s) + H2(g)
Given the following:
2H20(g) + 2CO(g) 2H2(g) + 2CO2(g) K = 2.56
3FeO(s) + 3CO(g) 3Fe(s) + 3CO2(g) K = .301
I assume the K values shown are for the equations AS WRITTEN.
2H20(g) + 2CO(g) ==> 2H2(g) + 2CO2(g) K = 2.56
3FeO(s) + 3CO(g) ==> 3Fe(s) + 3CO2(g) K = .301
You want this. Fe(s) + H2O(g) ==> FeO(s) + H2(g)
Leave eqn 1 as is with K = 2.56
Reverse eqn 2 so K for eqn 2 is 1/0.301 = 3.22
2H20(g) + 2CO(g) ==> 2H2(g) + 2CO2(g) K = 2.56
3Fe(s) + 3CO2(g) ==> 3FeO(s) + 3CO(g) K = 3.22
Next leave eqn 1 as is.
Divide the coefficients of eqn 2 by 3 and multiply by 2, The K = (3.22)^2/3
2H20(g) + 2CO(g) ==> 2H2(g) + 2CO2(g) K = 2.56
2Fe(s) + 2CO2(g) ==> 2FeO(s) + 2CO(g) K = (3.22)^2/3
Now add the last two equations.
2Fe(s) + 2H2O(g) ==> 2FeO(s) + 2H2(g) K = 2.56 x (3.22)^2/3 but this is for double what you want so take the square root of the result.
Post your work if you get stuck.
So the answer is 2.36?
To find the equilibrium constant, K, for the reaction Fe(S) + H2O(g) = FeO(s) + H2(g), we can use the equilibrium constant expression known as the "law of mass action." The general form of the law of mass action is as follows:
aA + bB = cC + dD
The equilibrium constant expression for this reaction is given by:
K = ([C]^c * [D]^d) / ([A]^a * [B]^b)
where [A], [B], [C], and [D] represent the concentrations of the corresponding species at equilibrium, and a, b, c, and d represent the stoichiometric coefficients of the balanced equation.
In this case, the balanced equation for the reaction is:
Fe(S) + H2O(g) = FeO(s) + H2(g)
The stoichiometric coefficients are:
a = 1 (for Fe(S))
b = 1 (for H2O(g))
c = 1 (for FeO(s))
d = 1 (for H2(g))
Thus, the equilibrium constant expression can be written as:
K = ([FeO]^1 * [H2]^1) / ([Fe]^1 * [H2O]^1)
However, since K is not given directly in the given information, we need additional information or data to determine the value of K for this specific reaction.