Hot copper turnings can be used as an oxygen getter for inert gas supplies by slowly passing the gas over the turnings at 600K

2Cu+1/2O₂(g)=Cu₂O(s); deltaG=-124.944 kj

how many molecules of oxygen are left in 1-litre of a gas supply after equilibrium has been reached?plz explain...why we calculate K_p using deltaG ...why not K_c....

To determine how many molecules of oxygen remain in a 1-liter gas supply after equilibrium has been reached, we first need to understand the relationship between the Gibbs free energy change (ΔG) and the equilibrium constant (K). The reaction you're looking at is:

2Cu + 1/2O₂ (g) ⇌ Cu₂O (s)

The given ΔG for this reaction is -124.944 kJ. This negative value indicates that the reaction is spontaneous in the forward direction under standard conditions. Now, let's delve into how we can calculate the equilibrium constant (K) and why we use K_p instead of K_c.

Understanding Equilibrium Constants

The equilibrium constant (K) can be expressed in two forms: K_c for concentrations and K_p for partial pressures. The relationship between ΔG and K is given by the equation:

ΔG = ΔG_° + RT ln(Q)

At equilibrium, ΔG = 0, and the equation simplifies to:

0 = ΔG_° + RT ln(K)

From this, we can rearrange to find:

ln(K) = -ΔG_° / RT

Exponentiating both sides gives us:

K = e^(-ΔG_° / RT)

Why K_p Instead of K_c?

In your reaction, oxygen is in the gaseous state, and we are interested in the partial pressures of the gases involved. K_p is specifically used for reactions involving gases and is defined in terms of the partial pressures of the reactants and products. Since we are dealing with a gas (O₂), K_p is more appropriate. K_c, on the other hand, is used for reactions where concentrations (molarity) are more relevant, typically in solutions.

Calculating K_p and Remaining Oxygen Molecules

Now, let's calculate K_p using the given ΔG:

• R (universal gas constant) = 8.314 J/(mol·K) = 0.008314 kJ/(mol·K)
• T = 600 K

Substituting these values into the equation:

K_p = e^(-ΔG_° / RT) = e^(124.944 kJ / (0.008314 kJ/(mol·K) * 600 K))

Calculating the exponent:

ΔG_° = 124.944 kJ = 124944 J

Exponent = -124944 / (8.314 * 600) ≈ -25.06

Now, K_p = e^-25.06 ≈ 1.1 x 10^-11

Finding the Remaining Oxygen Molecules

At equilibrium, the expression for K_p for this reaction is:

K_p = P_O2^1/2 / (1) = P_O2^1/2

Let’s denote the partial pressure of O₂ at equilibrium as P_O2. Rearranging gives:

P_O2 = K_p²

Substituting the value of K_p:

P_O2 = (1.1 x 10^-11)² ≈ 1.21 x 10^-22 atm

Using the ideal gas law (PV = nRT), we can find the number of moles of O₂:

n = PV / RT

For a volume of 1 liter (0.001 m³):

n = (1.21 x 10^-22 atm) * (0.001 m³) / (0.0821 L·atm/(mol·K) * 600 K)

Calculating this gives:

n ≈ 2.46 x 10^-25 moles

To find the number of molecules, we multiply by Avogadro's number (6.022 x 10²³ molecules/mol):

Number of molecules = n * N_A ≈ (2.46 x 10^-25) * (6.022 x 10²³) ≈ 1.48 x 10^-1 molecules

This indicates that after equilibrium is reached, there are approximately 0.148 molecules of O₂ left in the gas supply, which is a very small amount, suggesting that the reaction heavily favors the formation of Cu₂O.