When one mole of CH3OH (l) forms from its elements at 1 atm and 25 degrees C, 238.6 kJ of heat are released, and 35.2 kJ are released when one mole of the gas condenses. How much heat is released when one mole of CH3OH (g) forms from its elements at these conditions?
C(graphite) + 2H2(g) + 1/2 O2(g) ==> CH3OH(l) dH = -283.6 kJ
CH3OH(g) ==> CH3OH(l) dH = -35.2 kJ.
Add equation 1 to the reverse of equation 2 to obtain elements to CH3OH(g), then add the dH for each reaction to obtain total rxn.
Post your work if you get stuck.
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To calculate the heat released when one mole of CH3OH (g) forms from its elements at the given conditions, we need to consider two steps: the formation of CH3OH (l) from its elements and the condensation of CH3OH (g) to form CH3OH (l).
Step 1: Formation of CH3OH (l) from elements
The given information states that when one mole of CH3OH (l) forms from its elements, 238.6 kJ of heat are released. This means that the reaction is exothermic, and the heat released corresponds to the enthalpy change (ΔH) of the reaction.
Step 2: Condensation of CH3OH (g) to CH3OH (l)
The given information also states that when one mole of the gas condenses, 35.2 kJ of heat are released. Since the condensation of a gas to a liquid is an exothermic process, this heat released also corresponds to the enthalpy change of the reaction.
To calculate the heat released when one mole of CH3OH (g) forms from its elements, we need to subtract the heat released during the condensation process from the heat released during the formation process.
Heat released during the formation of CH3OH (l) = 238.6 kJ/mol
Heat released during the condensation of CH3OH (g) = 35.2 kJ/mol
Heat released when one mole of CH3OH (g) forms = Heat released during the formation - Heat released during condensation
= 238.6 kJ/mol - 35.2 kJ/mol
= 203.4 kJ/mol
Therefore, when one mole of CH3OH (g) forms from its elements at 1 atm and 25 degrees C, approximately 203.4 kJ of heat are released.