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.

ΔH0 = -75.0 kJ 
Pressure, P = 43.0 atm 
= 43 atm * 101325 Pa/1 atm 
= 4356975 Pa 
= 4356975 N / m2 
Volume change, ΔV = 2.0 L - 5.0 L 
= -3.0 L 
= -3.0 *10-3 m3 
Work done, w = - PΔV 
= - (4356975 N / m2) (-3.0 *10-3 m3) 
= 13070 N m 
= 13070 J 
= 13.07 kJ 
We know that at consatnt pressure q = ΔH0 = -75.0 kJ 
Also ΔE = q + w 
= - 75.0 kJ + 13.07 kJ 
= -61.93 kJ