[alanjz]

When MgO reacts with H2O at 25 degrees C and 1 atm, the volume change is 04.6 mL*mol^-1.
MgO(s) + H2O (l) --> Mg(OH)2 (s)
What is the value of delta H - delta E for this reaction?
(a) -4.7x10^-1 J*mol^-1
(b) -4.7x10^2 J*mol^-1
(c) 4.7x10^2 J*mol^-1
(d) 4.7x10^-1 J*mol^-1

I'm pretty sure it has something to do with work and -PdeltaV but I don't know how to proceed. Like I tried: -(1)(-4.6*10^-3).

[Stovn0611]

Do you know of any formulas relating delta H, delta E, and work?

[alanjz]

isn't it dE=dH+w so dH-dE=-w=-(-PdV)=(1)(4.6x10^-3)?

[Stovn0611]

is (1 atm)(4.6*10^-3) measured in the correct units though? You might need to do a conversion since 1 L*atm =/= 1 J

[alanjz]

ah so I need to multiply that by 101.3 thanks a lot!