When trinitrotoluene TNT) detonates according to the reaction 2 C₇H₅N₃O₆ (s) $\rightarrow$ 3 N₂(g) +5 H₂(g) + 12 CO(g) + 2 C(s), the enthalpy change at 25 C and 1 atm is -1475 kJ while the volume increases by about 489 L.What is the internal energy change (assume only P-V work is done)?

Question

Quizplus · Accepted Answer

The internal energy change (ΔU) can be calculated using the formula ΔU = ΔH - PΔV, where ΔH is the enthalpy change, P is the pressure (in this case, 1 atm converted to kJ by multiplying by 101.3 J to get the value in kJ/L), and ΔV is the volume change. Given ΔH = -1475 kJ and ΔV = 489 L, converting P from atm to kJ/L (1 atm ≈ 101.3 J/L = 0.1013 kJ/L), PΔV = 0.1013 kJ/L * 489 L = 49.5367 kJ. Therefore, ΔU = -1475 kJ - 49.5367 kJ ≈ -1525 kJ.

When Trinitrotoluene TNT) Detonates According to the Reaction 2 →\rightarrow→

When Trinitrotoluene TNT) Detonates According to the Reaction
2 $\rightarrow$