Calculate ?_rG for the evaporation of methanol at 25°C under nonstandard conditions when $\mathrm { PCH } _ { 3 } \mathrm { OH }$ = 10.0 mbar
$\quad$ $\quad$ $\quad$ CH₃OH(l) ? CH₃OH(g)

Question

Calculate ?_rG for the evaporation of methanol at 25°C under nonstandard conditions when $\mathrm { PCH } _ { 3 } \mathrm { OH }$ = 10.0 mbar $\quad$$\quad$$\quad$CH₃OH(l)? CH₃OH(g)

Quizplus · Accepted Answer

The Gibbs free energy change for the evaporation of methanol can be calculated using the equation: $\Delta G = \Delta G^\circ + RT \ln(Q)$. At 25°C, $\Delta G^\circ$ for methanol is approximately +3.2 kJ/mol. Given that the pressure of methanol vapor is 10.0 mbar, which is less than the standard pressure of 1 bar, the reaction quotient $Q$ is less than 1, making $\ln(Q)$ negative. This results in a negative $\Delta G$, specifically -7.1 kJ/mol.

Calculate ?rG for the Evaporation of Methanol at 25°C Under PCH3OH\mathrm { PCH } _ { 3 } \mathrm { OH }PCH3​OH

Calculate ?_rG for the Evaporation of Methanol at 25°C Under $\mathrm { PCH } _ { 3 } \mathrm { OH }$