8.000 moles of H₂(g)react with 4.000 mol of O₂(g)to form 8.000 mol of H₂O(l)at 25°C and a constant pressure of 1.00 atm.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 ΔE° = +576.06 kJ. B) ΔH° = +546.4 kJ and ΔE° = +516.8 kJ. C) ΔH° = -546.4 kJ and ΔE° = -516.8 kJ. D) ΔH° = -546.4 kJ and ΔE° = -576.0 kJ.

Question

Quizplus · Accepted Answer

ΔH° represents the change in enthalpy, and ΔE° represents the change in internal energy. Since 546.4 kJ of heat are released, ΔH° is negative, indicating an exothermic reaction. The relationship between ΔH and ΔE is given by ΔH = ΔE + PΔV. Given that PΔV = -29.60 kJ, and ΔH° = -546.4 kJ, solving for ΔE° gives ΔE° = ΔH° - PΔV = -546.4 kJ - (-29.6 kJ) = -516.8 kJ.

8000 Moles of H2(g)react with 4

8000 Moles of H₂(g)react with 4