1. When you look at a molecular orbital energy level diagram of a row two homonuclear diatomic molecule such as O2, N2, or F2, you will see two pairs of molecular orbitals on the same energy level (meaning that they are "degenerate"). These are π2p bonding and π2p* antibonding molecular orbitals, so answer B is correct.
2. This one is a bit involved. First consider the relationship ΔE = q + w. At constant pressure, the work term is expressed as "pressure-volume" work and is w = –PΔV where ΔV = Vfinal �?Vinitial. Therefore, the relationship under constant pressure conditions is ΔE = q –PΔV. Under these conditions, q = ΔH, the enthalpy change of the system.
OK, if ΔE is positive and ΔH (q) is negative, this can only happen if ΔV is negative (P will always be a positive number):
ΔE = q �?nbsp; PΔV
(+) (�? (+)(�?
Since ΔV = Vfinal �?Vinitial = negative value, Vfinal must be less than Vinitial, which would mean the gas contracts from volume Vinitial to volume Vfinal. That narrows us down to answer B or D.
Since q is negative, that means the system is losing heat (–q means heat is lost by the system = exothermic process, +q means heat is gained by the system = endothermic process). That would be answer D. I hope!
Steve
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