For the formation of ammonia, the equilibrium
constant is known to be 5.2 × 10−5
at
25◦C. After analysis, it is determined that
[N2] = 3.1 M and [H2] = 0.8 M, both at equilibrium.
How many grams of ammonia are in
the 10 L reaction vessel at equilibium? Use
the equilibrium equation
N2(g) + 3 H2(g) ⇀↽ 2 NH3(g)
Answer in units of g
45
To solve this problem, we first need to calculate the equilibrium concentrations of NH3. The balanced equation tells us that the ratio of moles of N2 to moles of NH3 is 1:2.
Given that [N2] = 3.1 M at equilibrium, the concentration of NH3 can be calculated as follows:
[NH3] = (1/2) * [N2]
[NH3] = (1/2) * 3.1 M
[NH3] = 1.55 M
Now, we can use the concentration of NH3 to calculate the moles of NH3 in the 10 L reaction vessel:
Moles NH3 = [NH3] * Volume
Moles NH3 = 1.55 M * 10 L
Moles NH3 = 15.5 mol
Finally, we can use the molar mass of NH3 (17.03 g/mol) to calculate the mass of NH3:
Mass NH3 = Moles NH3 * Molar mass NH3
Mass NH3 = 15.5 mol * 17.03 g/mol
Mass NH3 = 263.86 g
Therefore, there are approximately 263.86 grams of NH3 in the 10 L reaction vessel at equilibrium.
To find the number of grams of ammonia at equilibrium, we need to use the given equilibrium constant and concentrations of nitrogen and hydrogen.
1. Calculate the concentration of ammonia using the equilibrium constant:
- The equilibrium constant expression for the given reaction is: Kc = [NH3]^2 / ([N2] * [H2]^3).
- Substitute the given values: Kc = (x^2) / (3.1 * (0.8^3)).
- Since the equilibrium concentrations of nitrogen and hydrogen were not given, we'll consider x as the change in concentration of ammonia from 0 M to equilibrium. Therefore, the concentration of ammonia at equilibrium is x M.
2. Solve for x:
- Rearrange the equilibrium constant expression: (x^2) / (3.1 * (0.8^3)) = 5.2 × 10^(-5).
- Multiply both sides by (3.1 * (0.8^3)) to isolate x^2: x^2 = 5.2 × 10^(-5) * (3.1 * (0.8^3)).
- Calculate the right-hand side of the equation: 5.2 × 10^(-5) * (3.1 * (0.8^3)) ≈ 4.088 × 10^(-6).
- Take the square root of both sides: x ≈ √(4.088 × 10^(-6)).
- Calculate the value of x: x ≈ 0.002022 M.
3. Calculate the number of moles of ammonia in the reaction vessel:
- The equilibrium concentration of ammonia is 0.002022 M in a 10 L reaction vessel.
- Moles of ammonia = concentration * volume = 0.002022 M * 10 L.
- Moles of ammonia ≈ 0.02022 moles.
4. Convert moles to grams:
- The molar mass of ammonia (NH3) is 17.03 g/mol.
- Grams of ammonia = moles * molar mass = 0.02022 moles * 17.03 g/mol.
Therefore, there are approximately 0.3446 grams of ammonia in the 10 L reaction vessel at equilibrium.
.........N2(g) + 3 H2(g) ⇀↽> 2 NH3(g)
E.......3.1M......0.8M..........x
K = 5.2E-5 = (NH3O^2/(N2)(H2)^3
Plug the equilibrium numbers into Keq and solve for (NH3) in mol/L.
Convert to mols in 10 L and convert that to grams.