# 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] =3M and [H2]=0.8M, both at equilibrium. How many grams of ammonia are in the 10L reaction vessel at equilibium? Use the equilibrium equation

N2(g) + 3H2(g) ⇀ ↽ 2NH3(g)

Answer in units of g

## 1.2 g NH3

## Are you ready for a chemical balancing act? Let's clown around with some numbers!

To find the grams of ammonia at equilibrium in the reaction vessel, we need to use the equilibrium constant and the concentrations of N2 and H2.

The equilibrium constant, Kc, is given as 5.2 × 10^−5. This means that at equilibrium, the product of the concentrations of NH3 squared (raised to the power of 2) divided by the product of the concentrations of N2 and H2 cubed (raised to the power of 3) will be equal to 5.2 × 10^−5.

So, let's set up the equation using the given concentrations:

(2NH3)^2 / ([N2] * [H2]^3) = 5.2 × 10^−5

Plug in the values:

(2x)^2 / (3 * 0.8^3) = 5.2 × 10^−5

Simplifying,

4x^2 / (1.92) = 5.2 × 10^−5

Now, let's solve for x by cross-multiplying:

4x^2 = 5.2 × 10^−5 * 1.92

Divide both sides by 4:

x^2 = (5.2 × 10^−5 * 1.92) / 4

Taking the square root of both sides:

x = sqrt((5.2 × 10^−5 * 1.92) / 4)

Now, plug in the numbers and calculate the value of x. And voila! You will find the grams of ammonia in the 10L reaction vessel at equilibrium.

## To find the amount of ammonia in the 10L reaction vessel at equilibrium, we can use the equilibrium constant expression and the given concentrations of nitrogen gas (N2) and hydrogen gas (H2).

First, let's set up the equilibrium constant expression:

Kc = [NH3]^2 / ([N2] * [H2]^3)

Given:

Kc = 5.2 × 10^-5

[N2] = 3M

[H2] = 0.8M

Now, substitute the given values into the equilibrium constant expression:

5.2 × 10^-5 = [NH3]^2 / (3 * 0.8^3)

Simplify the equation:

5.2 × 10^-5 = [NH3]^2 / (1.92 * 10^-2)

Cross-multiply to solve for [NH3]^2:

[NH3]^2 = 5.2 × 10^-5 * (1.92 * 10^-2)

[NH3]^2 = 9.98 * 10^-7

Take the square root of both sides to solve for [NH3]:

[NH3] ≈ √(9.98 * 10^-7)

[NH3] ≈ 9.99 * 10^-4 M

Now, we can calculate the amount of ammonia in moles in the 10L reaction vessel:

moles of NH3 = [NH3] * volume

≈ (9.99 * 10^-4 M) * (10 L)

≈ 9.99 * 10^-3 moles

To convert moles to grams, we need to multiply by the molar mass of ammonia (NH3):

molar mass of NH3 = 14.01 g/mol

grams of NH3 = (9.99 * 10^-3 moles) * (14.01 g/mol)

≈ 0.1399 g

Therefore, there are approximately 0.1399 grams of ammonia in the 10L reaction vessel at equilibrium.

## To find the number of grams of ammonia in the 10L reaction vessel at equilibrium, we need to use the given concentrations of nitrogen gas ([N2] = 3M) and hydrogen gas ([H2] = 0.8M), as well as the equilibrium constant (K = 5.2 × 10^−5) for the reaction:

N2(g) + 3H2(g) ⇀ ↽ 2NH3(g)

First, we need to convert the concentrations of N2 and H2 to moles. Remember that molarity (M) is defined as moles of solute per liter of solution.

For nitrogen gas (N2):

Number of moles of N2 = [N2] × volume of solution (in liters)

Number of moles of N2 = 3M × 10L = 30 moles

For hydrogen gas (H2):

Number of moles of H2 = [H2] × volume of solution (in liters)

Number of moles of H2 = 0.8M × 10L = 8 moles

Next, we need to use the stoichiometry of the reaction to determine the number of moles of ammonia (NH3) that can be formed. From the balanced equation, we see that 1 mole of N2 reacts to form 2 moles of NH3, and 3 moles of H2 react to form 2 moles of NH3.

Since the ratio of N2 to NH3 is 1:2, and the ratio of H2 to NH3 is 3:2, we need to consider which reagent is the limiting reagent. The limiting reagent is the one that produces the least amount of product based on the given amounts.

To determine the limiting reagent, we compare the stoichiometric coefficients of N2 and H2 with the number of moles of each reactant. The limiting reagent is the one with the smaller ratio. In this case, 30 moles of N2 and 8 moles of H2 are available. Since the ratio of N2 to H2 is 1:3, we can see that N2 is the limiting reagent.

Now, let's calculate the number of moles of NH3 formed from the limiting reagent:

Number of moles of NH3 = (moles of limiting reagent × stoichiometric coefficient of NH3) / stoichiometric coefficient of limiting reagent

Number of moles of NH3 = (30 moles × 2 moles of NH3) / 1 mole of N2

Number of moles of NH3 = 60 moles

Finally, we need to convert the moles of NH3 to grams. The molar mass of NH3 is 17.03 g/mol.

Mass of NH3 = number of moles of NH3 × molar mass of NH3

Mass of NH3 = 60 moles × 17.03 g/mol

Mass of NH3 = 1021.8 g

Therefore, there are approximately 1021.8 grams of ammonia in the 10L reaction vessel at equilibrium.

## K = (NH3)^2/(N2)(H2)^3

You know N2 and H2, substitute with Kc and find NH3 in M = mols/L.

Then to find mols in 10L, that is

mols = M x L

Then mol = grams/molar mass. You know molar mass and mols, solve for grams.