#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.
#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")#
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Answer 2

The change in internal energy of combustion is approximately -936.89 kJ/mol for the given conditions.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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