Answer: A
w= −P∆V =−(1.00 atm)(8.00 liters) = −8.00 L ⋅atm
Because the gas was expanding, w is negative (work was being done by the system).
- The molar heat of sublimation for molecular iodine is 62.30 kJ/mol at 25°C and 1.00 atm.
Calculate the ∆H in (J ⋅mole –1) for the reaction
I 22 ^sgh"I_ i R=8 314. J/mole K:
A. 8 314 298.62 30.:B. 62.30 −(8.314)(298)
C. 62.30 + (8.314)(298)D. 62.30(1000) + 1 ⋅(8.314)(298)
E. none of the above are correctAnswer: D
∆n= moles of gaseous product −moles of gaseous reactants
= 1 −0 = 1Use the equation ∆H =∆E +∆n ⋅RT
.
.
62 30
1(^1000) 1 8 314 298
mole
kJ
kJ
=+#:::J J/mole K K
9 __iiC
Part II: Specific Topics