Consider the evaporation of methanol at 25.0 °C :
CH3OH(l ) ¡ CH3...

Question

Consider the evaporation of methanol at 25.0 °C :
CH3OH(l ) ¡ CH3OH( g)
a. Find ΔG°
r at 25.0 °C.

Accepted Answer

Hey everyone today, we're being asked to calculate the standard Delta G. of the reaction for the evaporation of ethanol at 25°C. Now to find the delta G of the reaction in this case. Since we're given no other values, we actually need to utilize the standard free energy of formation for both the product and the reactant. Now the standard delta G. Of formation, our experimental values that are calculated and can be found in your textbook Now the delta G. Of formation for the reactant, which is the liquid ethanol. Delta G. Formation of liquid is negative 1 74. kg joules per mole. And the standard delta G. Of formation for the gas form of ethanol negative 67.9 killer jewels from all. And we can utilize this in the equation or the formula that the standard delta G of reaction is equal to the delta G. Of formation of the products G. Of formation of the product minus the delta G of formation of the reactant. We have to factor in how many moles of everything we have. But since it is a 1-1 mole ratio from liquid to gas, it's fairly simple. We'll have one mole times negative 1 67.9 kg jules Permal. Since that is the delta G. Of formation for the product the gasses form of ethanol minus one more of liquid ethanol, which has a delta G. Of formation of negative 1 74.8 kg joules per mole. And Adding this up, we get that the Delta T. of reaction for this reaction, the evaporation of ethanol at 25°C is 6.9 killer jewels. Permal. I hope this helps, and I look forward to seeing you on the next one.

Consider the evaporation of methanol at 25.0 °C : CH3OH(l ) ¡ CH3OH( g) a. Find ΔG° r at 25.0 °C.

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