Question

Find ΔE° for the reaction below if the process is carried out at a constant pressure of 1.00 atm andΔV (the volume change) = -24.5 L. (1 L ∙ atm = 101 J)

2 CO(g) + O2 (g) → 2 CO2(g) ΔH° = -566. kJ

a. +2.47 kJ
b. -568 kJ
c. -2.47 kJ
d. -564 kJ

Answer 1

d. -564 kJ

Step-by-step explanation:

It is possible to find ΔE° using the formula:

ΔH° = ΔE° + PΔV

For the reaction:

2CO(g) + O₂(g) → 2CO₂(g) ΔH°= -566kJ

As P is 1,00 atm and ΔV is -24,5L:

-566 kJ = ΔE° + 1,00atm×-24,5L×
`(0,101kJ)/(1atmL)`

-566 kJ = ΔE° - 2,47 kJ

ΔE° = -563,53 kJ

The ΔE° is:

d. -564 kJ

I hope it helps!