Question #ca68e

1 Answer
Jan 29, 2015

Answer:

#DeltaG_("rxn")# wil be equal to #"+2.72 kJ/mol"#.

Explanation:

(Long answer, but I had no choice...)

Once again, before doing anything else, you must determine what the concentration of the dissolved carbon dioxide will be

#CO_(2(g)) rightleftharpoons CO_(2(aq))#

This is done by using Henry's law, which implies that

#P_("gas") = k_H * C_("gas")#, where

#P_("gas")# - the partial pressure of the gas;
#k_H# - Henry's constant - it has a specific value for every gas and it's temperature-dependent;
#C_("gas")# - the concentration of the dissolved gas.

Since you're at #"25"^@"C"#, the value of Henry's constant is listed at #"29.41 L atm/mol"#, which means that (again, convert mbar to atm)

#C_(CO_2) = P_(CO_2)/k_H = "0.00691 atm"/"29.41 L atm/mol" = "0.000235 mol/L"#

You can now use this value to determine the given reaction's equilibrium constant

#CO_(2(aq)) + H_2O_((l)) rightleftharpoons H_((aq))^(+) + HCO_(3(aq))^(-)#

#K_c = ([H^(+)] * [HCO_3^(-)])/([CO_2]) = (0.0000398 * 0.0117)/(0.000235) = 0.00198#

(I've converted all the given concentrations into #"mol/L"#).

Now you add the extra #H^(+)# to the equilibrium. The new concentration of #H^(+)# will be

#H_("new")^(+) = H^(+) + H_("added")^(+) = 39.8mumol + 79.6 mumol = 119.4mumol#

Since the concentration of one of the species involved in the equilibrium changes, you can calculate the reaction quotient for that change

#Q_c = ([H_("new")^(+)] * [HCO_3^(-)])/([CO_2]) = (0.000119 * 0.0117)/(0.000235) = 0.00592#

The first observation you can make is that #Q_c# is biger than #K_c#, which is in accordance with the fact that we now have more products than reactants; as a result the equilibrium will shift to the left and favor the formation of more reactants.

The Gibbs free energy change for a system that has a reaction quotient #Q_c# is

#DeltaG_("rxn") = DeltaG^@ + RT*ln(Q_c)#

Since at equilibrium #DeltaG^@ = -RT * ln(K_c)#, the above equation becomes

#DeltaG_("rxn") = RT * ln(Q_c/K_c)#

Therefore,

#DeltaG_("rxn") = "8.314 J/mol K" * "298.15 K" * ln(0.00592/0.00198)#

#DeltaG_("rxn") = "2715 J/mol" = "+2.72 kJ/mol"#

A positive value for #DeltaG_("rxn")# means that the forward reaction will not be spontaneous; however, the reverse reaction will be spontaneous.