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

d. -564 kJ

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×

-566 kJ = ΔE° - 2,47 kJ

ΔE° = -563,53 kJ

The ΔE° is:

d. -564 kJ

I hope it helps!

it's c (engineer's do their work with mechanical stuff not on any living being)

the correct answer would be b) volume versus temperature, because they are also directly proportional.