The relation between vapour pressure , enthalpy of vapourisation and temperature is
ln (88/ 39) = DeltaH / 8.314 (1 / 318 - 1 / 298)
0.814 = DeltaH / 8.314 (2.11 X 10^-4 )
DeltaH = -32.07 kJ
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-20.16 KJ of heat are released by the reaction of 25.0 g of Na2O2.
Explanation:
Given:
mass of Na2O2 = 25 grams
atomic mass of Na2O2 = 78 gram/mole
number of mole =
=
=0. 32 moles
The balanced equation for the reaction:
2 Na2O2(s) + 2 H2O(l) → 4 NaOH(aq) + O2(g) ∆Hο = −126 kJ
It can be seen that 126 KJ of energy is released when 2 moles of Na2O2 undergoes reaction.
similarly 0.3 moles of Na2O2 on reaction would give:
=
x =
= -20.16 KJ
Thus, - 20.16 KJ of energy will be released.