A sample of iron(II) sulfate was heated in an evacuated container to 920 K, where the following reactions occurred.

Reaction (a): 2 FeSO4(s)--> Fe2O3(s) + SO3(g) + SO2(g)
Reaction (b): SO3(g)--> SO2(g) + 1/2 O2(g)
After equilibrium was reached, the total pressure was 0.836 atm and the partial pressure of oxygen was 0.0275 atm. Calculate Kp for each of these reactions

I'm having the most trouble with finding the partial pressures.

To find the partial pressures in this problem, you need to use the given total pressure and the mole ratios of the reactants and products.

Let's start by looking at reaction (b): SO3(g) ⟶ SO2(g) + 1/2 O2(g)

From the stoichiometry of this equation, we can see that for every 1 mole of SO3 that reacts, you produce 1 mole of SO2 and 0.5 moles of O2.

Now, let's assume that the initial moles of SO3 is 'x'. According to the equation, the amount of SO2 and O2 produced will be 'x' and '0.5x', respectively.

Since we have the partial pressure of oxygen (0.0275 atm), we can equate it to the moles of O2 multiplied by the constant R and the temperature in Kelvin:

0.0275 atm = (0.5x) * R * 920 K

Here, R is the ideal gas constant (0.0821 L·atm/(mol·K)).

Now, let's solve this equation for 'x':

0.0275 atm = 0.5x * 0.0821 L·atm/(mol·K) * 920 K

Simplifying the equation:

0.0275 atm = 37.954x

Dividing both sides by 37.954:

x = 0.000725 mol

Now that we know the moles of SO3, we can find the moles of SO2:

moles of SO2 = x = 0.000725 mol

To find the partial pressure of SO2, we will use the ideal gas law equation:

PV = nRT, where P is the partial pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

Since we are given the total pressure (0.836 atm) and the volume is constant, we can rearrange the equation to solve for the partial pressure:

P = nRT / V

P(SO2) = (0.000725 mol) * (0.0821 L·atm/(mol·K)) * (920 K) / V

Given that V and R are constant, we can simplify the equation further:

P(SO2) = 0.0602 atm

Similarly, by using the same method, you can calculate the partial pressure of SO3 and find it to be 0.0602 atm.

Now that we have the partial pressures of SO2 and SO3, we can go back to the equation for reaction (a): 2FeSO4(s) ⟶ Fe2O3(s) + SO3(g) + SO2(g)

From the stoichiometry, we can see that for every 2 moles of FeSO4 that react, you produce 1 mole of Fe2O3, 1 mole of SO3, and 1 mole of SO2.

Thus, the moles of FeSO4 that reacted would be twice the moles of SO3 or SO2:

moles of FeSO4 reacted = 2 * (0.000725 mol) = 0.00145 mol

To find the partial pressure of Fe2O3, we will use the same method as before:

P(Fe2O3) = (0.00145 mol) * (0.0821 L·atm/(mol·K)) * (920 K) / V = 0.1204 atm

Therefore, we have the partial pressures as follows:
P(SO3) = 0.0602 atm
P(SO2) = 0.0602 atm
P(Fe2O3) = 0.1204 atm

Now, to calculate Kp for each reaction, we need to use the formula:

Kp = (P(SO3) * P(SO2)) / (P(Fe2O3))

Substituting the values we obtained:

Kp (reaction a) = (0.0602 atm * 0.0602 atm) / 0.1204 atm = 0.0301

Kp (reaction b) is not necessary to calculate because it only involves a single reaction step.

So, the value of Kp for reaction (a) is 0.0301.

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