At constant pressure for the reaction shown below,what can be said about PΔV and ΔE? N₂(g)+ 3 H₂(g)→ 2 NH₃(g)ΔH° = - 92.2 kJ

Question

Quizplus · Accepted Answer

At constant pressure, PΔV (change in pressure times change in volume) is negative since the volume of the reactants decreases as they form products. This means that the system is doing work on the surroundings, so PΔV < 0.
ΔE (change in internal energy) is equal to q (heat added to or removed from the system) plus w (work done on or by the system). Since the reaction is exothermic (ΔH° = -92.2 kJ), q is negative (heat is released). Since PΔV is also negative (work is done by the system), w is negative as well. Therefore, ΔE = q + w is negative, meaning that ΔE < -92.2 kJ.

Therefore, the best choice is C: PΔV < 0 and ΔE > -92.2 kJ.

At Constant Pressure for the Reaction Shown Below,what Can Be