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During the formation of $\ce{Al2O3}$ from 5.4 grams of $\ce{Al}$ and enough oxygen, the temperature of 2 kg of water climbs by 20 degrees. What’s the enthalpy of formation of $\ce{Al2O3}$ (per mole) ?

What I tried:

The balanced reaction:

$$\ce{4Al + 3 O2 -> 2 Al2O3}$$

5.4 grams of $\ce{Al}$ are 0.2 moles. If 2 kg of water climbed by 20 degrees then 40 000 calories were invested during the reaction. To get one mole of $\ce{Al2O3}$ we need 2 moles of $\ce{Al}$ , therefore the enthalpy of formation (per mole) of $\ce{Al2O3}$ is 400 000 calories. However, the correct answer is −400 000.

Can someone please explain the negative sign?

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$\ce{Al2O3}$, upon the formation of its lattice, releases heat. This heats up water. The increase in the temperature is in water, not $\ce{Al2O3}$. We usually refer to the water as the surroundings and $\ce{Al2O3}$ as the thermodynamic system.

A thermodynamic system is the content of a macroscopic volume in space, along with its walls and surroundings; it undergoes thermodynamic processes according to the principles of thermodynamics. Wikipedia

So, upon the formation of lattice (among with other, mostly, endothermic reactions) a lot of heat is released and that's what you have estimated. Since $\ce{Al2O3}$ is the system, it loses heat. Thus, $\Delta H_f$ is negative.

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I guess you are confused by sign convention of physics and chemistry. In chemistry, aluminium oxide is the system, and the reaction releases energy, and that energy is obviously coming out of the system. Since it is a loss of energy for the system it is referred as negative sign. But in physics the sign convention is just the opposite.

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  • $\begingroup$ Are you sure the sign convention is opposite in physics for heat released? Because if you're referring to the first law of thermodynamics, then the sign conventions for the work done (W) and not the heat exchanged (Q). Can you please confirm? Thanks! $\endgroup$ – Gaurang Tandon Feb 27 '18 at 2:28

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