The reaction b/t H2 and O2 yield water vapor and has a change of heat =-484kJ.Ho much PV work is done and what is the value of the change of E in kJ for the reaction of 0.50 mol of H2 with 0.25 mol of O2 amd atmospheric pressure of 1.00 , the volume change is -5.61.

To calculate the PV work done in a chemical reaction, you can use the equation:

PV work = -PΔV

Where:
P: pressure (in this case, atmospheric pressure = 1.00 atm)
ΔV: change in volume (in this case, -5.61 L)

Substituting the values:

PV work = -(1.00 atm) * (-5.61 L)
PV work = 5.61 atm L

Now let's calculate the change in internal energy (ΔE) using the equation:

ΔE = q + w

Where:
q: heat exchanged in the reaction (given as -484 kJ)
w: work done by the system (PV work)

Substituting the values:

ΔE = (-484 kJ) + (5.61 atm L)
ΔE = -484 kJ + 5.61 atm L

Note that atm L conversion to kJ is required. To convert atm L to J, multiply by 101.3 J/atm. Since 1 kJ = 1000 J, you need to multiply by (101.3 J/atm) * (1 kJ/1000 J) = 0.1013 kJ/atm.

ΔE = -484 kJ + (5.61 atm L) * (0.1013 kJ/atm)
ΔE = -484 kJ + 0.5694 kJ
ΔE = -483.4306 kJ

Therefore, the change in internal energy (ΔE) in kJ for the reaction of 0.50 mol of H2 with 0.25 mol of O2 at atmospheric pressure is -483.4306 kJ. The PV work done is 5.61 atm L.

-483kJ

=-484kJ-[1.0atm[-5.60L]]
=-484kJ-[5.60Latm[101j/Latm]]
=-484kJ+565.5J
=-483kJ