Question

In: Chemistry

Find ΔE° for the reaction below if the process is carried out at a constant pressure...

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

Select one:

A. +2.47 kJ

B. -568 kJ

C. -2.47 kJ

D. -564 kJ

Solutions

Expert Solution

Given ΔH° = -566. kJ

Work done , W = P x ΔV

                       = 1 atm x (-24.5 L )

                      = -24.5 L atm

                      = -24.5x101 J              Since 1 L atm = 101 J

                     = -2.47x103 J

                     = -2.47 kJ

From first law of thermodynamics , ΔH° = ΔE° + W

                                                   ΔE° = ΔH° - W

                                                           = -566 kJ - (-2.47 kJ)

                                                           = -564 kJ

Therefore option (D) is correct

queen_honey_blossom answered 2 years ago
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