Question

In a fuel cell, methanol is used as a fuel and oxygen gas is used as an oxidizer. At 298 K, the Gibb's free energy of formation for $CH_3OH(l)$, $H_2O(l)$ and $CO_2(g)$ are -168, -237, -394 kJ $mol^{-1}$ respectively. If the efficiency of the fuel cell is 80%, the enthalpy of combustion of methanol is -x kJ/mol. The value of x is _____.

Answer 1

875

Answer 2

The combustion reaction of methanol is: $CH_3OH(l) + \frac{3}{2}O_2(g) \longrightarrow CO_2(g) + 2H_2O(l)$

1. Calculate the standard Gibbs free energy change ($\Delta G^\circ$) for the reaction:

The standard Gibbs free energy of reaction is calculated using the standard Gibbs free energies of formation ($\Delta G^\circ_f$) of products and reactants:

$\Delta G^\circ_{reaction} = \sum (\text{products}) - \sum (\text{reactants})$

$\Delta G^\circ_{reaction} = [1 \times \Delta G^\circ_f(CO_2(g)) + 2 \times \Delta G^\circ_f(H_2O(l))] - [1 \times \Delta G^\circ_f(CH_3OH(l)) + \frac{3}{2} \times \Delta G^\circ_f(O_2(g))]$

Given values:

$\Delta G^\circ_f(CH_3OH(l)) = -168 \text{ kJ } mol^{-1}$ $\Delta G^\circ_f(H_2O(l)) = -237 \text{ kJ } mol^{-1}$ $\Delta G^\circ_f(CO_2(g)) = -394 \text{ kJ } mol^{-1}$ $\Delta G^\circ_f(O_2(g)) = 0 \text{ kJ } mol^{-1}$ (element in its standard state)

Substitute the values into the equation:

$\Delta G^\circ_{reaction} = [1 \times (-394 \text{ kJ } mol^{-1}) + 2 \times (-237 \text{ kJ } mol^{-1})] - [1 \times (-168 \text{ kJ } mol^{-1}) + \frac{3}{2} \times 0 \text{ kJ } mol^{-1}]$

$\Delta G^\circ_{reaction} = [-394 - 474] - [-168]$

$\Delta G^\circ_{reaction} = -868 - (-168)$

$\Delta G^\circ_{reaction} = -868 + 168$

$\Delta G^\circ_{reaction} = -700 \text{ kJ } mol^{-1}$

2. Use the efficiency formula for a fuel cell:

The efficiency ($\eta$) of a fuel cell is given by the ratio of the maximum useful work (Gibbs free energy change) to the total energy released (enthalpy change):

$\eta = \frac{\Delta G^\circ}{\Delta H^\circ}$ (Here, both $\Delta G^\circ$ and $\Delta H^\circ$ are negative, so their ratio is positive, representing efficiency).

Given efficiency $\eta = 80\% = 0.80$.

3. Calculate the enthalpy of combustion ($\Delta H^\circ$):

We have $\Delta G^\circ = -700 \text{ kJ } mol^{-1}$ and $\eta = 0.80$.

Rearrange the efficiency formula to solve for $\Delta H^\circ$:

$\Delta H^\circ = \frac{\Delta G^\circ}{\eta}$

$\Delta H^\circ = \frac{-700 \text{ kJ } mol^{-1}}{0.80}$

$\Delta H^\circ = -875 \text{ kJ } mol^{-1}$

4. Determine the value of x:

The enthalpy of combustion of methanol is given as -x kJ/mol.

So, $\Delta H^\circ = -x \text{ kJ } mol^{-1}$.

Comparing this with our calculated value:

$-x = -875$

$x = 875$