The reversible reaction n2(g) + 3h2(g)<-> produces ammonia, which is a fertilizer. At quilibrium, a 1-L flask contains 0.15 mol H2, 0.25 mol N2, and 0.10 mol NH3. Calculate K eq for a reverse reaction?
Please HELPP!
I applaud your writing a question with capital letters to begin sentences and placing comma's, periods, and questions marks appropriately. However, you need to sharpen those skills in writing equations since n2 and h2 have different meanings in lower case versus upper case.
N2(g) + 3H2(g) ==> 2NH3(g)
Write the Keq expression.
Kc = (N2)(H2)^3/(NH3)^2
All you need to do is to calculate the molarity of each component and substitute into the Kc expression.
M = moles/L; therefore,
(H2) = moles/L = 0.15/1 L = 0.15 M
(N2) = 0.25 mol/L etc.
To calculate the equilibrium constant (K_eq) for the reverse reaction, you need to make use of the stoichiometric coefficients of the balanced equation. The balanced equation for the reaction is:
N2(g) + 3H2(g) ⇌ 2NH3(g)
Given the initial concentrations of H2, N2, and NH3 in the flask, we can calculate K_eq as follows:
K_eq = ([NH3]^2) / ([N2] * [H2]^3)
Where:
[NH3] = concentration of NH3
[N2] = concentration of N2
[H2] = concentration of H2
From the information provided, we have:
[NH3] = 0.10 mol / 1 L = 0.10 M
[N2] = 0.25 mol / 1 L = 0.25 M
[H2] = 0.15 mol / 1 L = 0.15 M
Substituting these values into the equation, we get:
K_eq = (0.10^2) / (0.25 * 0.15^3)
Calculating this expression gives:
K_eq = 0.04 / (0.0375 * 0.003375)
K_eq = 0.04 / 1.27 x 10^-5
K_eq = 3.15 x 10^3
Therefore, the equilibrium constant (K_eq) for the reverse reaction is 3.15 x 10^3.
To calculate the equilibrium constant (Keq) for the reverse reaction, you need to use the concentrations of the reactants and products at equilibrium. In this case, you're given the initial amounts and need to determine the equilibrium concentrations.
Step 1: Write the balanced equation for the reverse reaction:
N2(g) + 3H2(g) ⇌ 2NH3(g)
Step 2: Determine the concentrations at equilibrium:
Given:
Initial concentration of H2 = 0.15 mol/L
Initial concentration of N2 = 0.25 mol/L
Initial concentration of NH3 = 0.10 mol/L
Note that the stoichiometry of the balanced equation tells us that the concentration of NH3 is twice the concentration of N2.
Let 'x' be the change in concentration of N2 (and NH3) at equilibrium.
The concentration at equilibrium for N2 will be: 0.25 - x
The concentration at equilibrium for NH3 will be: 0.10 + 2x
Step 3: Substitute the equilibrium concentrations into the equilibrium expression:
Keq = [NH3]^2 / ([N2] * [H2]^3)
Since the reverse reaction is taking place, we use the equilibrium concentrations for NH3, N2, and H2 as determined in Step 2.
Keq = (0.10 + 2x)^2 / ((0.25 - x) * (0.15)^3)
Step 4: Solve for Keq:
Since the reaction is at equilibrium, the value of x will make the concentrations of NH3 and N2 reach equilibrium. You can set up an equation to solve for x.
0.10 + 2x = 0.25 - x
3x = 0.15
x = 0.05
Substituting this value of x into the expression for Keq:
Keq = (0.10 + 2(0.05))^2 / ((0.25 - 0.05) * (0.15)^3)
Keq = (0.10 + 0.10)^2 / (0.20 * 0.003375)
Keq = 0.20^2 / 0.000675
Keq = 0.04 / 0.000675
Keq ≈ 59.26
So, the equilibrium constant (Keq) for the reverse reaction is approximately 59.26.