# Question ca68e

Jan 29, 2015

$\Delta {G}_{\text{rxn}}$ wil be equal to $\text{+2.72 kJ/mol}$.

#### Explanation:

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

$C {O}_{2 \left(g\right)} r i g h t \le f t h a r p \infty n s C {O}_{2 \left(a q\right)}$

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

${P}_{\text{gas") = k_H * C_("gas}}$, where

${P}_{\text{gas}}$ - the partial pressure of the gas;
${k}_{H}$ - Henry's constant - it has a specific value for every gas and it's temperature-dependent;
${C}_{\text{gas}}$ - the concentration of the dissolved gas.

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

${C}_{C {O}_{2}} = {P}_{C {O}_{2}} / {k}_{H} = \text{0.00691 atm"/"29.41 L atm/mol" = "0.000235 mol/L}$

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

$C {O}_{2 \left(a q\right)} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {H}_{\left(a q\right)}^{+} + H C {O}_{3 \left(a q\right)}^{-}$

${K}_{c} = \frac{\left[{H}^{+}\right] \cdot \left[H C {O}_{3}^{-}\right]}{\left[C {O}_{2}\right]} = \frac{0.0000398 \cdot 0.0117}{0.000235} = 0.00198$

(I've converted all the given concentrations into $\text{mol/L}$).

Now you add the extra ${H}^{+}$ to the equilibrium. The new concentration of ${H}^{+}$ will be

${H}_{\text{new")^(+) = H^(+) + H_("added}}^{+} = 39.8 \mu m o l + 79.6 \mu m o l = 119.4 \mu m o l$

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

${Q}_{c} = \frac{\left[{H}_{\text{new}}^{+}\right] \cdot \left[H C {O}_{3}^{-}\right]}{\left[C {O}_{2}\right]} = \frac{0.000119 \cdot 0.0117}{0.000235} = 0.00592$

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

$\Delta {G}_{\text{rxn}} = \Delta {G}^{\circ} + R T \cdot \ln \left({Q}_{c}\right)$

Since at equilibrium $\Delta {G}^{\circ} = - R T \cdot \ln \left({K}_{c}\right)$, the above equation becomes

$\Delta {G}_{\text{rxn}} = R T \cdot \ln \left({Q}_{c} / {K}_{c}\right)$

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 $\Delta {G}_{\text{rxn}}$ means that the forward reaction will not be spontaneous; however, the reverse reaction will be spontaneous.