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-127 KJmol-1
it can never be spontanious
It is not spontaneous.
3600 K
10,267 kJ are needed
H2(g) + S(s) —> H2S + 20.6 kJ
-127 KJmol-1
just add them together and you get 147kj
The enthalpy of combustion is determined by calorimetry.
[from wikipedia] The standard enthalpy of formation"standard heat of formation" of a compound is the change of enthalpy that accompanies the formation of 1 mole of a substance in its standard state from its constituent elements in their standard states (the most stable form of the element at 1 bar of pressure and the specified temperature, usually 298.15 K or 25 degrees Celsius). Its symbol is ΔHfO.
it can never be spontanious
It is not spontaneous.
400 K
it is never spontaneous
I2(s) --> I2(g); dH=62.4kJ/mol; dS=0.145kJ/mol. The reaction will favor the product at this temperature. Your entropy is positive and your enthalpy is also positive, so this reaction will not be spontaneous at all temperatures. But at room temperature, which is 298K, it will be spontaneous and proceed left to right. (this is the sublimation of iodine)
C(s)+O2(g) yields CO2(g)+393.5kJ
It is spontaneous