Question

Calculate the work and energy change when 67.11g Cu_2O is oxidized at a constant pressure of 1.00 bar and a constant temperature of 25 degrees Celsius.

a. ΔW=0,ΔE=q
b. ΔW=q,ΔE=0
c. ΔW=−PΔV,ΔE=q
d. ΔW=q,ΔE=−PΔV

Answer 1

For the oxidation of Cu_2O at constant pressure and temperature, the work done by the system is -PΔV, indicating energy lost by the system as the gas expands, and the change in energy is equal to the heat q added or removed from the system.

Step-by-step explanation:

The question is about calculating the work (W) and change in energy (ΔE) when 67.11g of Cu2O is oxidized at a constant pressure and temperature. When the reaction occurs under constant pressure, with a system that can perform PV work, the work done (W) by the system is equal to the product of the external pressure (P) and the change in volume (ΔV), given by the equation w = -PΔV. This work is done against atmospheric pressure by a piston, which moves when gas is produced or consumed during the reaction.

The change in energy (ΔE) of the system is the heat (q) gained or lost during the process. Since we're dealing with a closed system at constant pressure, the heat exchanged is equal to the enthalpy change (ΔH), however, more information would be needed to calculate heat or enthalpy change specifically for the described reaction. Therefore, among the options provided, the correct relationship would be ΔW = -PΔV and ΔE = q, corresponding to option (c).