A student ran the following reaction in the laboratory at 651 K:
2 NH3(g) N2(g) + 3 H2(g)
When she introduced 8.75E-2 moles of NH3(g) into a 1.00 Liter container, she found the equilibrium concentration of H2(g) to be 0.119 M.
Calculate the equilibrium constant, Kc, she obtained for this reaction.
See your other post above. Same problem. Different numbers.
To calculate the equilibrium constant, Kc, for this reaction, we need to use the equilibrium concentrations of the reactants and products.
In this case, we are given the equilibrium concentration of H2(g) as 0.119 M. However, we also need to determine the equilibrium concentrations of N2(g) and NH3(g) to calculate Kc.
First, let's use the stoichiometry of the reaction to determine the equilibrium concentrations of N2(g) and NH3(g) based on the concentration of H2(g).
From the balanced equation, we see that the stoichiometric ratio between NH3(g) and H2(g) is 2:3. This means that for every 3 moles of H2(g) that react, 2 moles of NH3(g) will also react. Therefore, the concentration of NH3(g) should be (2/3) * 0.119 M = 0.0793 M.
Similarly, the stoichiometric ratio between N2(g) and H2(g) is 1:3. So, the concentration of N2(g) can be calculated as (1/3) * 0.119 M = 0.0397 M.
Now that we have the equilibrium concentrations of all species, we can use these values to calculate Kc.
The expression for Kc for this reaction is:
Kc = ([N2] * [H2]^3) / [NH3]^2
Plugging in the values, we get:
Kc = (0.0397 * (0.119)^3) / (0.0793)^2
Calculating this expression, we find:
Kc = 1.04
Therefore, the student obtained an equilibrium constant, Kc, of approximately 1.04 for this reaction at 651 K.