Problem: The decomposition of isopropanol to form acetone and hydrogen gas (reaction shown below) has an equilibrium constant of 0.44 at 452 K. Assume ΔH° and ΔS° are temperature-independent.(CH3)2CHOH (g) ⇋ (CH3)2CO (g) + H2 (g)isopropanol              acetoneUseful data (at 298 K):(CH3)2CHOH (g)    ΔH°f = -261.1 kJ/mol(CH3)2CO (g)          ΔH°f = -218.5 kJ/molWill the reaction proceed spontaneously as written at 452 K when the partial pressures of each of the three species are 0.25 atm? Justify your answer with appropriate calculations.

FREE Expert Solution

We’re being asked to determine the if the reaction proceed spontaneously as written at 452 K when the partial pressures of each of the three species are 0.25 atm.


(CH3)2CHOH(g) ⇋ (CH3)2CO(g) + H2(g)



Recall that if:

• ΔG < 0 or ΔG = (–); the reaction is spontaneous

ΔG = 0; the reaction is at equilibrium

• ΔG > 0 or ΔG = (+); the reaction is non-spontaneous


Recall that ΔG˚rxn and K are related to each other:


ΔG°rxn=-RTlnK


When the system is not at equilibrium the equation of Gibbs Free energy changes to: 

ΔGrxn=ΔG°rxn+RTlnQ



For this problem, we need to do the following steps:

Step 1: Calculate for ΔG˚rxn at 452 K

Step 2: Calculate for Q

Step 3: Calculate ΔGrxn.



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Problem Details

The decomposition of isopropanol to form acetone and hydrogen gas (reaction shown below) has an equilibrium constant of 0.44 at 452 K. Assume ΔH° and ΔS° are temperature-independent.

(CH3)2CHOH (g) ⇋ (CH3)2CO (g) + H2 (g)

isopropanol              acetone

Useful data (at 298 K):

(CH3)2CHOH (g)    ΔH°f = -261.1 kJ/mol

(CH3)2CO (g)          ΔH°f = -218.5 kJ/mol

Will the reaction proceed spontaneously as written at 452 K when the partial pressures of each of the three species are 0.25 atm? Justify your answer with appropriate calculations.

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