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$\ce{AlCl3}$ has a standard enthalpy of formation some high negative value at $298\ \mathrm K$ $(-705.63\ \mathrm{J/mol})$. It means it's an exothermic reaction i.e will release heat.

But my question is how can it release heat when it (aluminium) has not even started reacting at room temperature with chlorine.

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  • $\begingroup$ Shouldn't it be -705.63 kJ/mol, not J/mol? $\endgroup$ Sep 11, 2020 at 7:41
  • $\begingroup$ The value -705 doesn't matter and unit also I am asking if such a large energy is released i.e it's an exothermic reaction in 298k or room temperature,my question is how is it possible because at 298k Al and cl2 doesn't even react $\endgroup$
    – Debakant
    Sep 11, 2020 at 8:09
  • $\begingroup$ That was asked like a dozen times - it has nothing to do if it reacts, or even exists in room temp. It's just a standard value. $\endgroup$
    – Mithoron
    Sep 11, 2020 at 18:16

2 Answers 2

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The molar enthalpy of formation at room temperature is not dependent on the reaction effectively happening at the room temperature.

It follows the general principle of state variables, being related to the Hess's law, which refers to the fundamental law of energy conservation. The principle says, that the change of enthalpy ( or any other state variable ) does not depend on the way, how you reach the new state, but only on the initial and final states themselves. If it had not been so, we would have been able to easily construct a perpetuum mobile that would have been switching between 2 states along 2 different paths, producing energy from nothing.

Particularly applied to the $\ce{AlCl3(s)}$ formation enthalpy, it can be formally expressed as the sum of respective enthalpies of these formal steps (see the Hess's law):

Enthalpy of $\ce{Al}$ atomization: $\ce{Al(s) -> Al(g)}$
Enthalpy of $\ce{Cl}$ atomization: $\ce{3/2 Cl2(g) -> 3 Cl(g)}$
Enthalpy of reaction : $\ce{3 Cl(g) + Al(g) -> AlCl3(g)}$ ( more exactly $\ce{Al2Cl6(g)}$ )
Enthalpy of condensation : $\ce{AlCl3(g) -> AlCl3(s)}$

Another way of dealing with measurement of formation or reaction enthalpy, that are too slow at room temperature, is measuring it at high temperature.

Then $$\Delta H_{\mathrm{r},\mathrm{T}_1}= \sum_\mathrm{reactants} n_i \cdot C_i \cdot ( T_2 - T_1) + \Delta H_{\mathrm{r},\mathrm{T}_2} + \sum_\mathrm{products} n_i \cdot C_i \cdot ( T_1 - T_2) $$

$n_i$ are molar amounts.
$C_i$ are molar heat capacities.

That means the reaction enthalpy at room temperature = Enthalpy of heating up the reactants + reaction enthalpy at high temperature + enthalpy of cooling down the products.

All kinds of enthalpy chnages can be combined and sumed, i.e. enthalpy of formation, reaction, phase change, heat exchange etc.

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  • $\begingroup$ Yeah thanks. Let me sure am I right in my understanding. The enthalpy change is not measured directly by performing a particular reaction in this case Al +Cl2 but actually observing various other enthalpy changes that may be enthalpy of atomisation enthalpy of formation of Al etc . In other cases where so4²- etc involved then we can actually interpret the enthalpy of formation of so4²- in the reaction to calculate the enthalpy of formation of any other substance like FeSO4 etc. $\endgroup$
    – Debakant
    Sep 11, 2020 at 9:34
  • $\begingroup$ Yes. See also the answer update. All kinds of enthalpy chnages can be combined, i.e. enthalpy of formation, reaction, phase change, heat exchange etc. $\endgroup$
    – Poutnik
    Sep 11, 2020 at 9:50
  • $\begingroup$ @ Debakant. Forget about sulfate. You are right in saying that the enthalpy change is not measured directly, but is obtained by actually observing various enthalpy changes and combining them. First you measure the amount of heat $\ce{Q1}$ necessary to heat aluminum metal up to the temperature where it reacts with chlorine. Second you measure the amont of heat $\ce{Q2}$ produced by the reaction. Third you measure the amount of heat $\ce{Q3}$ produced by cooling the obtained $\ce{AlCl3}$ to room temperature. These three values are then combined to get your final enthalpy change. $\endgroup$
    – Maurice
    Sep 11, 2020 at 9:58
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You heat your calorimeter with Al and chlorine (electrically, recording the energy you put in) from 298K until the reaction starts.

You record the final temperature after the reaction.

After cooling the calorimeter down to 298K, you heat it (with the AlCl3 in it) until it is at the same temperature again, and record the energy needed.

Now you subtract the two energies, and (if you put 1mol Al and 3/2 mol Cl2 in the calorimeter) the difference will be, lo and behold, 706 kJ.

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