✔ 最佳答案
1.
First, you've wrong calculations on positive-negative values.
your first equation should be
enthalpy change
= 3 x -(std. ent. change of FORMATION of hydrazine) + 4 x (std. ent. change of FORMATION of ammonia)
= -3 x (+50.4) + 4 x (-46.3) = -290.1 kJ/mol
and your second equation = -290.1/3 = -96.7 kJ/mol
Secondly, notice what the question asks: "std. enthalpy change of decomposition".
Check if there's any definition for this term.
- If it says "per mole of reactant", then you'll need to divide the subtotal by no. of mole of reactant in equation
i.e. = -290.1 / 3 = -96.7 kJ/mol
- If it says "per mole of product", then you'll need to divide the subtotal by no. of mole of product in equation
i.e. = -290.1 / 4 = -72.3 kJ/mol
- If it says nothing, then you just calculate the values according to the equation, just like the "enthalpy change of reaction"
i.e. -290.1 kJ <<without "/mol" >>
- You can define this term by yourself, say "per mole of N2 formed"... then divide / multiply the values accordingly.
2.
notice that there's instantaneous dipole- induced dipole attraction (dispersion force) formed between ethanol and cyclohexane, albeit small.
energy is also given out when such attraction is formed.
HOWEVER, the energy given out is too small to compensate the energy required to overcome the hydrogen bond attraction, so....