For the conversion of glucose to G6P, what is #DeltaG_(rxn)^@#?

I got #DeltaG_(rxn) = "20.4 kJ/mol"# to three sig figs. We don't want #DeltaG^@#. We already have that!

The reaction was:

#"Glucose"(aq) + "Phosphate"(aq) rightleftharpoons "Glucose-6-phosphate"(aq) + "H"_2"O"(l)#, #DeltaG^@ = "13.8 kJ/mol"#

We know that:

• #["Glucose"] = "19.0 mmol/L"#
• #["Phosphate"] = "6.90 mmol/L"#
• #["Glucose-6-phosphate"] = "1.70 mmol/L"#
• #["H"_2"O"] = "55.5 mol/L"#

and we know that the NONSTANDARD change in Gibbs' free energy of reaction, #DeltaG_(rxn)#, is what we want...

That is related to the reaction quotient #Q# as follows:

#DeltaG = DeltaG^@ + RTlnQ#

where the standard state for #DeltaG^@# is defined to be at #25^@ "C"# and #"1 bar"#.

Among the concentrations, water is the tricky one here. The standard state of a solid or liquid is its molar density at #25^@ "C"# and #"1 bar"# pressure.

At #37.0^@ "C"#, its mass density is #"0.9933316 g/mL"#, so its molar density is:

#"993.3316 g"/"L" xx "1 mol"/"18.015 g" = "55.139 mol/L"#

In the mass action expression, we divide by the standard state units to make #Q_c# unitless, and for water we have:

#(["H"_2"O"])/(["H"_2"O"]^@) = "55.5 mol/L"/"55.139 mol/L" = 1.0065#,

which for example, is what we would insert as the current concentration value of water.

It is, as expected, close to #1#. This is why we don't include pure liquids and solids in mass action expressions.

Now, given all these concentrations, we presumably can calculate the reaction quotient of this reaction:

#Q_c = ((("0.00170 mol/L")/("1 mol/L")) cdot ("55.5 mol/L"/"55.139 mol/L"))/((("0.0190 mol/L")/("1 mol/L")) cdot ("0.00690 mol/L"/"1 mol/L"))#

#= 13.05#

Therefore, the change in Gibbs' free energy for this reaction at #37^@ "C"# (and presumably, #"1 bar"#) is, assuming #DeltaG^@# stays constant relative to #25^@ "C"#:

#color(blue)(DeltaG_(rxn)) = DeltaG_(rxn)^@ + RTlnQ_c#

#= "13.8 kJ/mol" + (8.314472 xx 10^(-3) "kJ/mol"cdot"K")(37+"273.15 K")ln(13.05)#

#=# #color(blue)"20.4 kJ/mol"#

Knowing that #DeltaG_(rxn) = 0# at equilibrium, which way does the reaction want to shift? (how does #Q# change to decrease #DeltaG_(rxn)#?)

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Did you determine that it shifted left? Since #DeltaG_(rxn)^@ < DeltaG_(rxn)#, #Q# must decrease to decrease #DeltaG_(rxn)#. That means the product concentration should decrease and thus, the reaction should shift to make more reactants.