Question

Find Δ E° for the reaction below if the process is carried out at a constant pressure of 1.00 atm and Δ V (the volume change) = -24.5 L. (1 L ∙ atm = 101 J) 2 CO(g) + O2 (g) → 2 CO2(g)ΔH° = -566. kJ

Answer 1

2 CO (g) + O₂ (g) → 2 CO₂ (g)

The given conditions are:

ΔH° = -566. kJ

Pressure = constant = 1.00 atm

ΔV = -24.5 L

ΔE° = ?

To solve this problem we will use this formula:

ΔH = ΔE° + PΔV

That's the formula that describes our process since it is a reaction at constant pressure.

ΔH = ΔE° + PΔV

ΔE° = ΔH - PΔV

ΔE° = (-566. kJ) - 1.00 atm * (-24.5 L)

ΔE° = - 566. kJ + 24.5 atm*L

ΔE° = - 566. kJ + 24.5 atm*L * 101 J/ (1 atm*L)

ΔE° = - 566. kJ + 2475 J

ΔE° = - 566. kJ + 2475 J * (1 kJ/1000 J)

ΔE° = - 566. kJ + 2.475 KJ

ΔE° = - 563.5 kJ