Question

An ideal gaseous reaction (which is a hypothetical gaseous reaction that conforms to the laws governing gas behavior) occurs at a constant pressure of 40.0 atm and releases 74.8 kJ of heat. Before the reaction, the volume of the system was 7.60 L . After the reaction, the volume of the system was 2.40 L . Calculate the total internal energy change, ΔE , in kilojoules

Answer 1

Since the process occurs at constant pressure, the heat transferred to the system equals its change in enthalpy. That means the amount of heat released equals the decrease of the system's enthalpy:
∆H = - 74.8 kJ

Enthalpy is defined in terms of internal energy U, pressure and volume:
H = U + P∙V
<=>
U = H + P∙V

So the changes are related as:
∆U = ∆H - ∆(P∙V)
Since P is constant you can rewrite
∆U = ∆H - P∙∆V = ∆H - P∙(V_final _ V_initial)

To compute the P∙V in kilojoules you need to covert pressure to kilopascals and volume to cubic meters.
Note that 1 kJ = 1 kPa∙m³

Hence,
∆U = - 74.8 kJ - 40*101.325 kPa * (2.4×10⁻³ m³ - 7.6×10⁻³ m³) = -53.72 kJ