Calculate $\Delta$G at 298 K for the reaction C₂H₅OH(l)$\rightarrow$ C₂H₅OH(g,0.0400 bar) Given $\Delta$G^$\omicron$ = 6.2 kJ at 298 K.

Question

Quizplus · Accepted Answer

$\Delta G = \Delta G^\circ + RT \ln P$, where $\Delta G^\circ = 6.2 \, \text{kJ}$, $R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1}$, $T = 298 \, \text{K}$, and $P = 0.0400 \, \text{bar} = 0.0400 \times 10^5 \, \text{Pa}$. Converting $\Delta G^\circ$ to J gives $6200 \, \text{J}$. Plugging in the values: $\Delta G = 6200 + (8.314 \times 298 \times \ln(0.0400 \times 10^5))$. The calculation yields a value close to $-1800 \, \text{J}$ or $-1.8 \, \text{kJ}$.

Calculate Δ\DeltaΔ G at 298 K for the Reaction C2H5OH(l) →\rightarrow→ C2H5OH(g,0 Δ\DeltaΔ

Calculate $\Delta$ G at 298 K for the Reaction
C₂H₅OH(l) $\rightarrow$ C₂H₅OH(g,0 $\Delta$