Find the warm that will certainly be released in kJ per gram once aluminium ($ceAl$) reacts via $ceFe2O3$ as follows: $$ce2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(l)$$

The information given is:

$Delta H = 12.40 space mkJ/mol$ (for $ceFe$)$Delta H = -822.2 oom mkJ/mol$ (for $ceFe2O3$)$Delta H = -1669.8 space mkJ/mol$ (for $ceAl2O3$)

If it asked me to find the heat released in kJ/mol that would certainly be easy:

$$Delta H^circ = amount u_iDelta H^circ_f ( mproducts) - amount u_iDelta H^circ_f ( mreactants)$$

I assumed that there is one mole and I discovered the mass in grams by multiplying the molar mass of each compound by one mole.

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Then I divided the $H^circ_f$"s by those amounts to obtain the new values "expressed in kJ per gram"

Then I provided the previous equation and I gained a solution around 10 yet the correct answer is 15.24 kJ/g.

The question is: is the method I provided wrong, or is tright here somepoint else I didn"t pay attention to?

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edited Sep 16 "19 at 12:19

Martin - マーチン♦
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asked Aug 11 "14 at 19:25

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When you converted the heats to kJ/g, you inadvertently "broke" the Hess"s law equation that you plugged them into.

Each of the enthalpies of development in that equation is multiplied by the variety of moles of the matching species, and so the devices of the enthalpy of formation are kJ/mol.

If you assume one mole of product and also convert that to grams, you then would certainly need to transform every one of the various other species to grams (2 moles of Al, two moles of Fe, etc).

It"s less complicated to just do the whole calculation in moles, then divide the last answer by grams. However before, you should specify which species is the gram basis. In other words, you are asking for the warm per gram, however per gram of what?

Whatever the answer is, then find the full warm output in terms of kJ/mol of that substance, then divide it by the molar mass of the substance (in g/mol).

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One means to stop these kinds of difficulties is to always explicitly incorporate the devices in any kind of calculation that you perform. In this instance, you would have actually conveniently checked out that the Hess"s Law equation would certainly not give you the correct devices.