Question

`1)` What is the change in internal energy of combustion in `"kJ/mol"` for a bomb calorimeter whose heat capacity is `"3024 J/"^@ "C"` that has raised in temperature by `1.2910^@ "C"` due to the combustion of `"0.1575 g Mg"`? Also help with one more?

`2)` If the change in enthalpy of reaction for `"2NO"(g) + "O"_2(g) -> 2"NO"_2(g)` is `-"114.6 kJ"`, then what amount of heat is produced when `"23.000 kg"` of `"NO"_2(g)` is generated?

Answer 1

I would have asked this in two separate questions...

`2)` `DeltaE_C = "602.5 kJ/mol"` was released from the bomb.

`3)` `DeltaH_(rxn) = 2.86 xx 10^4 "kJ"` for the scaled-up reaction.

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`2)` A bomb calorimeter has constant volume, so by the first law of thermodynamics,

`DeltaE = q + w`

`= q - cancel(PDeltaV)`

so the heat flow is equal to the change in internal energy in this scenario. You were given not the specific heat capacity, but the heat capacity... note the units are `"J/"^@ "C"`. That's on purpose, to make this simplified from a real scenario.

The amount of internal energy released from the bomb out to the water is:

`DeltaE = q = "3024 J/"cancel(""^@ "C") cdot (1.2910cancel(""^@ "C"))`

`= "3904 J" = "3.904 kJ"`

But the question wants you to report it in `"kJ/mol"`. Of what? Of the magnesium. That's what the mass is for. (And that is why the units are very important!)

Thus, the internal energy released from the magnesium is:

`color(blue)(DeltaE_C) = |-"3.904 kJ"/(0.1575 cancel"g Mg" xx "1 mol Mg"/(24.305 cancel"g Mg"))|`

`=` `color(blue)("602.5 kJ/mol")`

and the VALUE of `DeltaE_C` is `-"602.5 kJ/mol"`.

`3)` Well, your reaction gave `DeltaH` for the reaction as-written. That is unfortunately common. What should have been specified by the original question writer is:

`2"NO"(g) + "O"_2(g) -> 2"NO"_2(g)`

`DeltaH_(rxn) = -"114.6 kJ/"color(red)("2 mols NO"_2(g))`.

That then acts as a conversion factor. The heat PRODUCED asked for is the magnitude, not a negative value. So,

`color(blue)(DeltaH_(rxn)) = 2.30 xx 10^4 cancel("g NO"_2) xx cancel("1 mol NO"_2)/(46.0055 cancel("g NO"_2)) xx "+114.6 kJ"/(2 cancel("mol NO"_2))`

`= color(blue)(2.86 xx 10^4 "kJ")`