Amole 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.

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