# Calculate the standard free energy change for the below reaction at 25 degrees Celsius?

##
#CH_4"(g)"+2O_2"(g)"\toCO_2"(g)"+2H_2O"(l)"#

#\DeltaG_{f,H_2O"(g)"}^{o}=-237.2"kJ"/"mol"#

#\DeltaG_{f,CO_2"(g)"}^{o}=-394.4"kJ"/"mol"#

#\DeltaG_{f,CH_4"(g)"}^{o}=-50.8"kJ"/"mol"#

##### 1 Answer

I'll assume these are all gases (in a coffee-cup calorimeter), although water liquid is possible in a bomb calorimeter. As a note, you should be especially careful that the values you look at are correct.

Here I get

Two things that can give you trouble are:

- What phase is the substance at
#25^@ "C"# and#"1 atm"# ?#DeltaG_f^@# for#"H"_2"O"(g)# is NOT the same as for#"H"_2"O"(l)# . - Is the substance in its standard state already, i.e. is
#DeltaG_f^@# for the substance ZERO?

Due to the fact that the Gibbs' free energy

#DeltaG_(rxn)^@ = sum_P n_P DeltaG_(f,P)^@ - sum_R n_RDeltaG_(f,R)^@# where

#DeltaG_f^@# is the change in Gibbs' free energy of reaction at#25^@ "C"# and#"1 atm"# in#"kJ/mol"# , and#P# and#R# are products and reactants.#n# is the mols of substance.

*So, you take the sum over the products, sum over the reactants, and subtract the sums, making sure you get the phases correct, and the coefficients match up.*

Also, the **not** given the value, it is a good sign that you **already know it** "by heart".

**Important standard state examples** are:

#"H"_2(g), "C"(grap hite), "Al"(s), "P"_4(s), "S"_8(s)# (if you care, sulfur is actually orthorhombic in its standard state.) You should however definitely know that carbon graphite is the standard state of carbon, not diamond.)

The **main diatomics** to know here are

Therefore, for

#"CH"_4(g) + 2"O"_2(g) -> "CO"_2(g) + 2"H"_2"O"(g)#

we just have:

#color(blue)(DeltaG_(rxn)^@) = ["1 mol" cdot DeltaG_(f,CO_2(g))^@ + "2 mols" cdot DeltaG_(f,H_2O(g))^@] - ["1 mol" cdot DeltaG_(f,CH_4(g))^@ + "2 mols" cdot "0 kJ/mol O"_2(g)]#

#= ["1 mol" cdot -"394.4 kJ/mol CO"_2(g) + "2 mols" cdot -"237.2 kJ/mol H"_2"O"(g)] - ["1 mol" cdot -"50.8 kJ/mol CH"_4(g) + "2 mols" cdot "0 kJ/mol O"_2(g)]#

#=# #color(blue)(-"818 kJ")#