PCHEM/Partial Pressures Question

[Coffeepower]

[SOLVED] PCHEM/Partial Pressures Question

Problem Statement:
A vessel of volume 22.4dm[tex]^{}3[/tex] contains 2.0 mol H[tex]_{}2[/tex] and 1.0 mol of N[tex]_{}2[/tex] at 273.15K initially. All the H[tex]_{}2[/tex]reacted with sufficient N[tex]_{}2[/tex] to form NH[tex]_{}3[/tex]. Calculate the partial pressures and the total pressure of the final mixture.

Variables and Magnitudes:
V = 22.4dm[tex]^{}3[/tex]
T = 273.15K
n[tex]_{}H_{}2[/tex]= 2.0 mol
n[tex]_{}N_{}2[/tex]= 1.0 mol

First Attempt:
2H[tex]_{}2[/tex]+N[tex]_{}2[/tex][tex]\rightarrow[/tex]NH[tex]_{}3[/tex]
It takes 1.5 mol of H[tex]_{}2[/tex] to make 1 mol of NH[tex]_{}3[/tex] so I was thinking that I should say that:

p[tex]_{}H_{}2[/tex] = p[tex]_{}total[/tex]*X[tex]_{}H_{}2[/tex] = ?

We would need to calculate the total pressure first for this.

I need a place to start this problem. I just can't put my finger on it.

 

[Coffeepower]

After working on the problem some more; I found that there is a simple way to do the problem. Simply find the mole fractions with the number of moles given and calculate the total pressure with PV = nRT for 3 moles, and 22.4dm^3.

Not much more to it than that. The solution in the book matches up.

 

[Blueshift5]

To solve this problem, we can use the ideal gas law, which states that the total pressure of a gas mixture is equal to the sum of the partial pressures of each gas present. In this case, we have two gases present, H_{}2 and N_{}2, and we can use the ideal gas law to calculate their partial pressures.

First, we need to calculate the total number of moles of gas present in the vessel. Since we know the volume and temperature, we can use the ideal gas law to calculate the total number of moles, n_{}total:

n_{}total = (p_{}total * V) / (R * T)

Where p_{}total is the total pressure, V is the volume, R is the gas constant, and T is the temperature. We can rearrange this equation to solve for p_{}total:

p_{}total = (n_{}total * R * T) / V

Substituting the given values, we get:

p_{}total = (3.0 mol * 8.314 J/mol*K * 273.15 K) / 22.4 dm^{}3

p_{}total = 101.3 kPa

Now that we have the total pressure, we can use the mole ratios from the balanced chemical equation to calculate the partial pressures of each gas. From the equation, we know that for every 2 moles of H_{}2 that react, 1 mole of NH_{}3 is formed. This means that for every 1 mole of NH_{}3 formed, 2 moles of H_{}2 must have reacted. Therefore, the number of moles of H_{}2 remaining in the vessel is:

n_{}H_{}2 = 2.0 mol - (1.0 mol NH_{}3 / 2 mol H_{}2) = 1.5 mol H_{}2

Using the ideal gas law again, we can calculate the partial pressure of H_{}2:

p_{}H_{}2 = (n_{}H_{}2 * R * T) / V

Substituting the values, we get:

p_{}H_{}2 = (1.5 mol * 8.314 J/mol*K * 273.15 K) / 22.4 dm^{}3

p_{}H