When 6.000 moles of H2(g) reacts with 3.000 mol of O2(g) to form ...

[JMatthes]

Question 1

When 2.00 mol of benzene is vaporized at a constant pressure of 1.00 atm and at its normal boiling point of 80.1°C, 67.8 kJ are absorbed and PΔV for the vaporization process is equal to 5.80 kJ then
◦ ΔE = 62.0 kJ and ΔH = 67.8 kJ.
◦ ΔE = 67.8 kJ and ΔH = 62.0 kJ.
◦ ΔE = 67.8 kJ and ΔH = 73.6 kJ.
◦ ΔE = 73.6 kJ and ΔH = 67.8 kJ.

Question 2

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
◦ ΔH° = -409.8 kJ and ΔE° = -432.0 kJ.
◦ ΔH° = -409.8 kJ and ΔE° = -387.6 kJ.
◦ ΔH° = +409.8 kJ and ΔE° = +432.0 kJ.
◦ ΔH° = +409.8 kJ and ΔE° = +387.6 kJ.

[nyrave]

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[JMatthes]

Thanks for the timely response, appreciate it

[upturnedfurball]

Gracias!