Question

A mole of X reacts at a constant pressure of 43.0 atm via the reaction
X(g)+4Y(g)→2Z(g), ΔH∘=−75.0 kJ
Before the reaction, the volume of the gaseous mixture was 5.00 L. After the reaction, the volume was 2.00 L. Calculate the value of the total energy change, ΔE, in kilojoules.

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