Question

at 900 c the equilibrium constant Kc for the following reaction is 0.064. Fe2O3 + 3H2 = 2Fe = 3H2O. a mixture of 1 mol of Fe2O3 and 0.1 mol of H2 gas was added to a 2.0 l container and heated until the system reaches equilibrium. calculate the concentration of H2O and H2 at equilibrium

Answer 1

The concentration of H2O at equilibrium is 0.046 mol and the concentration of H2 at equilibrium is 0.054 mol.

Step-by-step explanation:

To calculate the concentration of H2O and H2 at equilibrium, we can use the given equilibrium constant (Kc) and the initial amounts of the reactants. According to the balanced equation:

Fe2O3 + 3H2 = 2Fe + 3H2O

The stoichiometry of the reaction tells us that for every 3 moles of H2 reacted, we will produce 3 moles of H2O. So, if 0.1 mol of H2 reacted, that means we will also produce 0.1 mol of H2O.

Now, let's use the equilibrium constant (Kc) to calculate the concentration of H2 and H2O at equilibrium:

Kc = [H2O]3/[H2]3

Since we know Kc = 0.064, we can rearrange the equation to:

0.064 = [H2O]3/[H2]3

Taking the cube root of both sides, we get:

0.064^(1/3) = [H2O]/[H2]

Solving for [H2O], we have:

[H2O] = 0.064^(1/3) * [H2]

Substituting the given value of [H2] = 0.1 mol into the equation, we get:

[H2O] = 0.064^(1/3) * 0.1 = 0.046 mol

So, the concentration of H2O at equilibrium is 0.046 mol. Since the stoichiometry of the reaction tells us that for every 3 mol of H2 reacted, we will produce 3 mol of H2O, the concentration of H2 at equilibrium will be 0.1 mol - 0.046 mol = 0.054 mol.