# How to determine the heat evolved or consumed by the reaction of 1.0 g SO2(g) with excess oxygen, with results from a Hess' Law Equation?

##### 1 Answer

**(A)**

Based off of the *state function* property of enthalpy, **Hess's Law** states that you can:

**Scale**a reaction stoichiometry#-># **scale**the enthalpy value**Reverse**a reaction#-># **Reverse the sign**of the enthalpy

And if you follow these two operations, you preserve the validity of your answer.

The reactions you have available are:

(1)#2"S"(s) + 3"O"_2(g) -> 2"SO"_3(g)# ,#Delta"H"_1 = -"790 kJ/mol"#

(2)#"S"(s) + "O"_2(g) -> "SO"_2(g)# ,#Delta"H"_2 = -"297 kJ/mol"# to achieve:

(3)#2"SO"_2(g) + "O"_2(g) -> 2"SO"_3(g)# ,#Delta"H"_"rxn" = ???#

These reactions are indeed error-free to begin with, so let's proceed.

- Notice again, that the sulfur solid is NOT in the final result. Thus, it must have disappeared in the middle of the reaction as an
**intermediate**. So... you must have to cancel that out. - In
**(2)**,#"SO"_2(g)# is a product, whereas in**(3)**, it's a reactant. Thus, you must reverse**(2)**. - To cancel out
#"S"(s)# now, you should double the coefficients in**(2)**.

Thus, your final operations become:

#cancel(2"S"(s)) + 3"O"_2(g) -> 2"SO"_3(g)#

#2("SO"_2(g) -> cancel("S"(s)) + "O"_2(g))#

#"--------------------------------------"#

#color(blue)(2"SO"_2(g) + "O"_2(g) -> 2"SO"_3(g))#

Convince yourself that the sulfur and oxygen are balanced!

Now, you can proceed to calculate the enthalpy:

#color(blue)(Delta"H"_"rxn") = Delta"H"_1 + (-2)Delta"H"_2#

#= -790 + (-2)(-297) "kJ/mol"#

#=# #color(blue)(-"196 kJ/mol")#

Also, be sure to clarify whether it's a standard reaction (i.e. the pressure according to STP, and

**(B)**

In this part, in knowing that you use "excess oxygen", you assume that **limiting reagent** (i.e. it is entirely consumed first, and the reaction ends after that point), and from there, utilize the following equation for heat flow at a constant pressure:

#\mathbf(Delta"H"_"rxn" = (q_"rxn")/"mols limiting reagent" = (q_"rxn")/(n_("SO"_2(g))))#

I assume the reaction conditions (i.e. temperature and pressure) of part **(A)** and **(B)** are the same.

Thus, we should get for the **heat flow** involved in this reaction:

#color(blue)(q_"rxn") = n_("SO"_2(g))Delta"H"_"rxn"#

#= stackrel(n_("SO"_2(g)))overbrace((1.0 cancel("g SO"_2))((cancel("1 mol SO"_2))/(64.063 cancel("g SO"_2))))stackrel(Delta"H"_"rxn")overbrace((-"196 kJ/"cancel"mol"))#

#= color(blue)(-"3.06 kJ")#

Since the sign of the heat flow is **negative**, heat is *produced* by the reaction and *leaves* the reaction. You can decide what the wording is for your answer.