8.000 moles of H₂(g)reacts with 4.000 mol of O₂(g)to form 8.000 mol of H₂O(l)at 25 °C and a constant pressure of 1.0133 bar.If 546.4 kJ of heat are released during this reaction,and PΔV is equal to -29.60 kJ,then: A)ΔH° = +546.4 kJ and ΔU = +576.06 kJ B)ΔH° = +546.4 kJ and ΔU = +516.8 kJ C)ΔH° = -546.4 kJ and ΔU = -516.8 kJ D)ΔH° = -546.4 kJ and ΔU = -576.0 kJ

Question

Quizplus · Accepted Answer

The correct answer is based on the understanding that ΔH (enthalpy change) and ΔU (internal energy change) are related by the equation ΔH = ΔU + PΔV. Given that 546.4 kJ of heat are released, ΔH° = -546.4 kJ (since release of heat is exothermic and thus negative). With PΔV = -29.60 kJ, to find ΔU, we rearrange the equation to ΔU = ΔH - PΔV = -546.4 kJ - (-29.60 kJ) = -516.8 kJ.

8000 Moles of H2(g)reacts with 4

8000 Moles of H₂(g)reacts with 4