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/molUnder standard conditions at 570 K, is this reaction spontaneous as written? Justify your answer with appropriate calculations.

FREE Expert Solution

We’re being asked to determine the if is this reaction spontaneous as written at 570 K.


(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


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


ln(K2K1)=-HrxnR[1T2-1T1]


We’re given the ΔH˚f  of each reactant and product:


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

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



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

Step 1: Calculate ΔH˚rxn.

Step 2: Calculate K at 570 K

Step 3: Calculate for ΔGrxn at 570 K



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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

Under standard conditions at 570 K, is this reaction spontaneous as written? Justify your answer with appropriate calculations.

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