An ideal gaseous reaction (which is a hypothetical gaseous reaction that conforms to the laws governing gas behavior) occurs at a constant pressure of 30.0 atm and releases 74.4 kj of heat. before the reaction, the volume of the system was 7.20 l . after the reaction, the volume of the system was 2.00 l . calculate the total internal energy change, δe, in kilojoules. express your answer with the appropriate units.

The total change in internal energy would simply be calculated using the formula:

ΔU = -P (V2 – V1) + ΔH

where ΔU is the change in internal energy; P is constant pressure = 30 atm = 3,039,750 Pa; V2 is final volume = 2 L = 0.002 m^3; V1 is initial volume = 7.20 L = 0.0072 m^3; while ΔH is the heat = -74,400 J (heat released so negative)

Therefore:

ΔU =-3,039,750 Pa * (0.002 m^3 - 0.0072 m^3) + (- 74,400 J)

ΔU = - 58,593.3 J = - 58.6 kJ

answer: enthalpy

explanation:

for a exothermic reaction, the energy is released i.e. the energy of the products is lower than the energy of the reactants.

= enthaply

= energy of products

= energy of reactants

as , thus comes out to be negative. negative value of enthalpy shows that the reaction is exothermic.