Answer to Question #257385 in Organic Chemistry for abdi

Question

(iii)  The enthalpy change (ΔH) for this reaction is –3509 kJ when one mole of pentane burns in oxygen.

The reaction is carried out at a pressure (p) of 100 kPa.

The change in volume (ΔV) is 0.21m³.

Calculate the change in internal energy (ΔU) for this reaction.

ΔH = ΔU + pΔV

Show your working.

(3)

Answer 1

ΔH = ΔU + pΔV

-3509= ∆U+100(0.21)

-3509= ∆U+21

∆U= -3509-21

∆U= -3530J