# How much heat will be released when 6.44 g of sulfur reacts with excess O2 according to the following reaction?

## ${\text{S"(s) + 3/2"O"_2(g) -> "SO}}_{3} \left(g\right)$, $\Delta {H}_{\text{rxn"^@ = -"395.7 kJ/mol}}$

Jan 2, 2016

STANDARD ENTHALPY OF REACTION

First, let's just check the standard enthalpy of formation for making one $\text{mol}$ of ${\text{SO}}_{3} \left(g\right)$.

$\Delta {\text{H}}_{f}^{\circ}$ for ${\text{SO}}_{3} \left(g\right)$ according to Principles of Chemistry: A Molecular Approach (Tro) is $- \text{395.7 kJ/mol}$. You made $\text{2 mol}$s of ${\text{SO}}_{3} \left(g\right)$. Multiply $\Delta {\text{H}}_{f}^{\circ}$ by $2$ and you have the enthalpy of reaction you were given.

We'll work with $\text{1 mol}$ of product to make it a bit simpler.

But wait a minute... that means in THIS case:

...the enthalpy of formation of the product fully contributes to the enthalpy of the reaction that makes $S {O}_{3} \left(g\right)$.

And if you look in your thermodynamic tables in your textbook appendix, you should see that oxygen as a diatomic substance (${\text{O}}_{2}$) in its elemental state only exists in one phase (a gas), and sulfur exists in its elemental state as a SOLID (not a gas; the gas form does have a nonzero enthalpy of formation).

So, based off of that, the reaction in standard conditions (defined under ${25}^{\circ} \text{C}$ and $\text{1 atm}$ for thermodynamic tables) with the phases is:

$\textcolor{g r e e n}{{\text{S"(s) + 3/2"O"_2(g) -> "SO}}_{3} \left(g\right)}$

$\textcolor{g r e e n}{\Delta {H}_{\text{rxn"^@ = -"395.7 kJ/mol}}}$

Now we just have to determine what the relationship between the heat released $q$ and the standard enthalpy of reaction $\Delta {\text{H"_"rxn}}^{\circ}$ is.

HEAT FLOW VS. ENTHALPY IN "GENERAL CHEMISTRY" CONDITIONS?

Something that probably no one told you in General Chemistry (even your professor or teacher) is that you are assuming constant-pressure conditions during your Thermodynamics unit.

I'm going to spare you the derivation, but you should know that when the pressure is kept constant, the heat flow in a reaction is equal to the enthalpy of reaction:

$\textcolor{g r e e n}{\Delta H = {q}_{p}}$

where ${q}_{p}$ is heat flow at a constant pressure. The units here specifically are typically $\text{kJ}$, though one can use $\text{J}$.

STANDARD ENTHALPY IS DEFINED ON A PER-MOL BASIS

If we note that the enthalpy of reaction given at standard conditions is on a per-mol basis (and I realize this may be a stretch, but bear with me):

The enthalpy of reaction is a normalization of the heat flow in the reaction.

What that means is that when you scale up any old reaction at constant $\text{1 atm}$ pressure and ${25}^{\circ} \text{C}$ so that exactly $\text{1 mol}$ of the important product is made (in this case, ${\text{SO}}_{3} \left(g\right)$), you have the same reaction as you see in standard thermodynamic tables you would find in your textbook appendices.

Lastly, only the limiting reagent is used up completely. Therefore, we can construct this equation:

\mathbf(DeltaH_"rxn"^@ = q_p/("mols of Limiting Reagent"))
with units of $\text{kJ/mol}$.

Did you notice how you have excess oxygen? That immediately tells you that the limiting reagent is sulfur.

SCALING YOUR REACTION TO THE STANDARD SCALE

So figuring out the scaling here:

("6.44" cancel"g S") / ("1" cancel"mol S" xx ("32.065" cancel"g S")/cancel("1 mol S")) xx 100%

~~ color(green)(20.08%)

So, what you have is $\text{0.2008 mol}$ of $\text{S}$ for the real-scale reaction, based on ONE $\text{mol}$ of product for the standard-scale reaction. Multiply $0.2008$ by the $\Delta {\text{H"_"rxn}}^{\circ}$, which is defined for ONE $\text{mol}$ of product ($- \text{395.7 kJ/mol}$), and you'll get the heat released to be:

$\textcolor{b l u e}{{q}_{p} = \text{79.47 kJ}}$

Be sure to mention that heat was released, if you write a positive answer. If you don't, write a negative answer and the release of heat is implied.