The oxidation of Cu₂O(s) to CuO(s) is an exothermic process: 2 Cu₂O(s) + O₂(g) → 4 CuO (s) ∆H = -292.0 kJ･mol⁻¹
Calculate the work done when 6.50 mol of Cu₂O(s) undergo oxidation at STP.

w = 7376.6 J

Explanation:

To calculate the work done in the system, we need first to calculate the number of moles of all compounds.

The following expression, that comes from the 1° law of thermodynamic, will help you to get the work:

w = ΔnRT

Where:

Δn: difference in the number of moles between products and reactants

R: constant universal of gases (8.314 J / mol K)

T: Temperature in Kelvin.

As we are doing this reaction in STP, then the Pressure is 1 atm, and temperature is 0 °C or 273 K.

Now, we already have the moles of Cu₂O, let's see, according to the balanced reaction, how many moles we should have of O₂ and CuO.

If 1 mole of O₂ reacts with 2 moles of Cu₂O then 6.5 moles of Cu₂O will be:

moles O₂ = 6.5 moles Cu₂O * (1 mole O₂/2 moles Cu₂O) = 3.25 moles O₂

If 2 moles of Cu₂O produces 4 moles of CuO, then 6.5 moles will be:

moles CuO = 6.5 moles Cu₂O*(4 moles CuO/2 moles Cu₂O) = 13 moles CuO

Now that we have the moles, let's calculate the value of Δn.

Δn = moles product - moles reactants

Δn = 13 - (6.5 + 3.25) = 3.25 moles

Now, we can calculate the work done in the system:

w = 3.25 * (8.314 * 273)

Hope this helps

fe 55.85 g/mol and h2o 18.02 g/mol.

sure im new too!

explanation: