Question

Consider the following reaction: `2Mg(s) + O_2(g) -> 2MgO(s)`, `DeltaH = -1204kJ`. Is this reaction exothermic or endothermic? How do you calculate the amount of heat transferred when 2.4 grams of Mg(s) reacts at constant pressure?

How many grams of MgO are produced during an enthalpy change of -96.0 kJ? How many kilojoules of heat are absorbed when 7.50 g of MgO(s) is decomposed into Mg(s) and `O_2(g)` at constant pressure?

Answer 1

This reaction as written is clearly `"exothermic.............."`

Explanation:

We has..............

`2Mg(s) + O_2(g) rarr 2MgO(s)` `DeltaH^@""_"rxn"=-1204*kJ*mol^-1` `(i)`

To make the arithmetic a bit easier we could write........

`Mg(s) + 1/2O_2(g) rarr MgO(s)` `DeltaH^@""_"rxn"=-602*kJ*mol^-1` `(ii)`

I am CLEARLY free to do this because `DeltaH^@` is quoted `"per mole of reaction as written.........."`

And so we combust a `(2.4*g)/(24.3*g*mol^-1)=0.0988*mol` quantity of metal.

And thus for the given molar quantity of magnesium, `DeltaH_"rxn"=0.0988*cancel(mol)xx-602*kJ*cancel(mol^-1)=-59.5*kJ`

So, in effect, we have treated enthalpy as a stoichiometric product of the reaction (which indeed it is!). And if `-96.0*kJ` energy are generated, with respect to `MgO`, this requires a molar quantity of `(-96.0*kJ)/(-602*kJ*mol^-1)=0.159*mol` of rxn `(ii)`, i.e. `0.159*molxx24.3*g*mol^-1=3.88*g` with respect to magnesium metal.

And when `7.5*g` metal are combusted, this should result in a heat output of ..........................

`(7.5*g)/(24.3*g*mol^-1)xx-602*kJ*mol^-1=-185.8*kJ`

Finally, you have proposed that a given mass of `MgO` is decomposed to magnesium metal, and oxygen gas. Clearly, the energy here must be supplied to the system for decomposition. Can you handle this yourself?