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.**

(CH_{3})_{2}CHOH_{(g)} ⇋ (CH_{3})_{2}CO_{(g)} + H_{2}_{(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:

$\overline{){\mathbf{\Delta G}}{{\mathbf{\xb0}}}_{{\mathbf{rxn}}}{\mathbf{=}}{\mathbf{-}}{\mathbf{RTlnK}}}$

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

$\overline{){{\mathbf{\Delta G}}}_{{\mathbf{rxn}}}{\mathbf{=}}{\mathbf{\Delta G}}{{\mathbf{\xb0}}}_{{\mathbf{rxn}}}{\mathbf{+}}{\mathbf{RTlnQ}}}$

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 ΔG_{rxn}.

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.

(CH_{3})_{2}CHOH (g) ⇋ (CH_{3})_{2}CO (g) + H_{2} (g)

isopropanol acetone

Useful data (at 298 K):

(CH_{3})_{2}CHOH (g) ΔH°_{f} = -261.1 kJ/mol

(CH_{3})_{2}CO (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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