# Question ad335

Apr 12, 2015

For this particular reaction, $\Delta {G}_{\text{rxn" = "+9.4 kJ/mol}}$.

$C {O}_{\left(g\right)} + 2 {H}_{2 \left(g\right)} r i g h t \le f t h a r p \infty n s C {H}_{3} O {H}_{\left(g\right)}$, ${K}_{p} = 2.26 \cdot {10}^{4}$

The equation you're going to use is

$\Delta {G}_{\text{rxn}} = \Delta {G}^{0} + R T \cdot \ln \left(Q\right)$, where

$\Delta {G}^{0}$ - the standard free energy change;
$R$ - the gas constant - ${\text{8.314 J mol"^(-1)"K}}^{- 1}$
$T$ - the temperature at which the reaction takes place - expressed in Kelvin;
$Q$ - the reaction quotient - expresses the relative amounts of rectans and products present at a particular point in a reaction.

For the given pressures, you need to see if the reaction is at equilibrium or not. To do this, calculate $\Delta {G}^{0}$ and $Q$ before solving for $\Delta {G}_{\text{rxn}}$.

So, at equilibrium, $\Delta {G}_{\text{rxn}} = 0$, which implies that

$\Delta {G}^{0} = - R T \cdot \ln \left({K}_{p}\right)$

DeltaG^0 = -8.314"J"/("mol" * cancel("K")) * (273.15 + 25)cancel("K") * ln(2.26 * 10^(4))

$\Delta {G}^{0} = - 24852.6 \text{J/mol" = -"24.85 kJ/mol}$

Now calculate $Q$

$Q = {P}_{C {H}_{3} O H} / \left({P}_{C O} \cdot {P}_{{H}_{2}}^{2}\right) = \frac{1.0}{1.0 \cdot {10}^{- 2} \cdot {\left(1.0 \cdot {10}^{- 2}\right)}^{2}} = {10}^{6}$

Because $Q > {K}_{p}$, the reaction will favor the reactants, i.e. you have more product present at these conditions than you'd have at equilibrium.

As a result, the forward reaction will no longer be favored, and you can expect $\Delta {G}_{\text{rxn}}$ to be positive $\to$ the reverse reaction will become spontaneous.

Plug these values into the main equation and solve for $\Delta {G}_{\text{rxn}}$

DeltaG_"rxn" = -"24.85 kJ/mol" + 8.314"J"/("mol" * cancel("K")) * (273.15 + 25)cancel("K") * ln(10^(6))

$\Delta {G}_{\text{rxn" = -"24.85 kJ/mol" + "34.25 kJ/mol}}$

DeltaG_"rxn" = color(green)(+"9.4 kJ/mol")#