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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

Respuesta :

jufsanabriasa

Answer:

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×[tex]\frac{0,101kJ}{1atmL}[/tex]

-566 kJ = ΔE° - 2,47 kJ

ΔE° = -563,53 kJ

The ΔE° is:

d. -564 kJ

I hope it helps!

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