energetics疑惑

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....

Why is the enthalpy of formation of N2 zero-.-? Does it mean that all elements in standard state do not need to be included in the calculation?

Refer to the question. It is NOT necessary to divide by 3.

However, if the question is "calculate the standard enthalpy change of the decomposition reaction FOR EACH MOLE OF N2H4", it is necessary to divide by 3.

2012-05-27 11:27:45 補充：
It is much more meaningful 'to explain that the hydrogen bonds are broken due to the movement of large cyclohexane molecules" than "to describe the formation of instantaneous dipole - induced dipole attraction between ethanol and cyclohexane molecules".

要除3的
standard enthalpy change of the decomposition
係１MOL分解

請參考2012 DSE

1.
3N2H4 → 4NH3 + N2

ΔHR of thereaction
= 4ΔHf[NH3]+ ΔHf[N2]- 3ΔHf[N2H4]
= 4 x (-46.3) + 0 - 3 x (+50.4)
= -336.4 kJ/mol

Note that "/mol" in the unit means that all coefficients in theequation are in mol.


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2.
In ethanol, the intermolecular forces between ethanol molecules are hydrogenbonds. In cyclohexane, theintermolecular forces between cyclohexane molecules are van der Waals' forces. When ethanol and cyclohexane are mixed, themigration of large cylcohexane molecules would break the hydrogen bonds betweenethanol molecules readily, and energy is absorbed when the hydrogen bonds arebroken. Therefore, mixing ethanol andcyclohexane is endothermic.

2012-05-27 11:13:21 補充：
Q.2
To answer this question, it is ONLY necessary to write the main point -- the breaking of hydrogen bonds between ethanol molecules. Also, it is better to explain why such hydrogen bonds are broken -- due to the migration of large cyclohexane molecules.