Question

A mole of X reacts at a constant pressure of 43.0 atm via the reaction

Answer 1

Below is the solution:

-62.0 kJ ΔH = -75.0 kJ, P = 43.0 atm 43atm x 101325 Pa/1 atm = 4356975 Pa or N/m2 ΔV = 2.0 L - 5.0 L = 3.0 L = -3.0 x 10-3 m3 w = -PΔV = -(4356975 N/m2) (-3.0 x 10-3 m3) = 13070 N m = 13070 J = 13.07 kJ At constant P, q = ΔH = -75.0 kJ & ΔE = q + w = - 75.0 kJ + 13.07 kJ = -61.93 kJ = -62.0 kJ

Answer 2

Answer:

Total energy change, ΔE = 61.929 kJ

Explanation:

Relationship between ΔH, ΔE and work done is given by first law of thermodynamics.

ΔE = ΔH - PΔV

Where,

ΔH = Change in enthalpy

ΔE = Change in internal energy

PΔV = Work done

Given,

ΔH = -75.0 kJ = -75000 J

P = 43.0 atm

ΔV = Final volume - initial volume

     = (2.00 - 5.00) = -3.00 L

PΔV = 43 × (-3.00) = -129 L atm

1 L atm = 101.325 J

-129 L atm = 129 × 101.325 = -13071 J

ΔE = ΔH - PΔV

     = -75000 - (-13071)

     = -75000 + 13071

     = 61929 J

Conversion from J to kilojoule

1 J = 10^-3 kJ

61929 J = $61929 \times 10^{-3}=61.929\ J$

Total energy change, ΔE = 61.929 kJ