When 6.000 moles of H₂(g)reacts with 3.000 mol of O₂(g)to form 6.000 mol of H₂O(l)at 25°C and a constant pressure of 1.00 atm.If 409.8 kJ of heat are released during this reaction,and PΔV is equal to -22.20 kJ,then A)ΔH° = +409.8 kJ and ΔE° = +432.0 kJ. B)ΔH° = +409.8 kJ and ΔE° = +387.6 kJ. C)ΔH° = -409.8 kJ and ΔE° = -387.6 kJ. D)ΔH° = -409.8 kJ and ΔE° = -432.0 kJ.

Question

Quizplus · Accepted Answer

Since heat is released during the reaction and PΔV is negative, the reaction is exothermic (-PΔV accounts for work done by the system). Therefore, ΔH° will be negative.

We can use the following equation:

ΔH° = ΔE° + PΔV

where ΔH° is the enthalpy change, ΔE° is the internal energy change, and PΔV is the pressure-volume work.

We know that ΔH° = -409.8 kJ and PΔV = -22.20 kJ (note that the negative sign indicates work done by the system). Therefore, we can solve for ΔE°:

ΔH° = ΔE° + PΔV

-409.8 kJ = ΔE° - 22.20 kJ

ΔE° = -387.6 kJ

So the answer is (C): ΔH° = -409.8 kJ and ΔE° = -387.6 kJ.

When 6000 Moles of H2(g)reacts with 3

When 6000 Moles of H₂(g)reacts with 3