# 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 ${\text{2NO"(g) + "O"_2(g) -> 2"NO}}_{2} \left(g\right)$ is $- \text{114.6 kJ}$, then what amount of heat is produced when $\text{23.000 kg}$ of ${\text{NO}}_{2} \left(g\right)$ is generated?

Jan 15, 2018

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

2) $\Delta {E}_{C} = \text{602.5 kJ/mol}$ was released from the bomb.

3) $\Delta {H}_{r x n} = 2.86 \times {10}^{4} \text{kJ}$ for the scaled-up reaction.

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

$\Delta E = q + w$

$= q - \cancel{P \Delta V}$

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 $\text{J/"^@ "C}$. 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"))

$= \text{3904 J" = "3.904 kJ}$

But the question wants you to report it in $\text{kJ/mol}$. 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"))|

$=$ $\textcolor{b l u e}{\text{602.5 kJ/mol}}$

and the VALUE of $\Delta {E}_{C}$ is $- \text{602.5 kJ/mol}$.

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

$2 {\text{NO"(g) + "O"_2(g) -> 2"NO}}_{2} \left(g\right)$

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))

$= \textcolor{b l u e}{2.86 \times {10}^{4} \text{kJ}}$