Question :

4 moles of A(g) is added to a fixed volume container and left until they arrive at the equilibrium status.

2A(g) <===> A2(g)

At 127 C total pressure of the equilibrium system is 8.314*10^5 Pa and the density of the mixture of gas is 10 kg/m^3 . If the relative atomic weight of A is 30 find the partial pressure of A2(g).


My first attempt towards solving this question is applying PM=dRT equation, where I took d as (30+x) (x is the relative atomic mass of A2(g) )
Then I got x=220
My next thought is finding moles of A2(g) and then obtaining the partial pressure of it.Then again I'm getting another unknown value for moles of A2(g) as follows;

2A(g) <===> A2(g)
initial(moles) 4 -
final (moles) (4-2y) y

How do I solve this?

To solve this question, you can use the principle of conservation of mass to determine the moles of A2(g) at equilibrium.

From the balanced equation, we know that 2 moles of A(g) react to form 1 mole of A2(g). Let's denote the moles of A2(g) at equilibrium as y.

Initially, we start with 4 moles of A(g). As the reaction proceeds, 2 moles of A(g) reacts to form y moles of A2(g). Therefore, the final number of moles of A(g) will be (4 - 2y) and the final number of moles of A2(g) will be y.

Now, we can use the ideal gas equation, PV = nRT, to relate the total pressure and the number of moles of a gas. In this case, the total pressure is given as 8.314 * 10^5 Pa and the temperature is 127°C, which we need to convert to Kelvin (K).

Let's plug in these values into the ideal gas equation:

Total pressure = partial pressure of A(g) + partial pressure of A2(g)
8.314 * 10^5 Pa = (4 - 2y) * R * T + y * R * T

Since both A(g) and A2(g) are at the same temperature, we can factor out R * T from both terms:

8.314 * 10^5 Pa = (4 - 2y + y) * R * T
8.314 * 10^5 Pa = (4 - y) * R * T

Now, let's consider the density of the mixture of gases. The density is given as 10 kg/m^3. Density is defined as mass/volume. The mass of the mixture can be calculated by multiplying the total moles of A(g) and the relative atomic weight of A.

Mass of the mixture = (4 - 2y) * atomic weight of A(g)

Since the density is known, we can calculate the volume of the mixture using the formula:

Volume = mass of the mixture / density
Volume = (4 - 2y) * atomic weight of A(g) / density

Now we have two equations:

1) 8.314 * 10^5 Pa = (4 - y) * R * T
2) Volume = (4 - 2y) * atomic weight of A(g) / density

Using these equations, you can solve for y, the moles of A2(g), and then calculate the partial pressure of A2(g) by substituting the value of y into the first equation.