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

(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:

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

Recall that **ΔH˚ _{rxn} and K** are related to each other:

$\overline{){\mathbf{ln}}\mathbf{\left(}\frac{{\mathbf{K}}_{\mathbf{2}}}{{\mathbf{K}}_{\mathbf{1}}}\mathbf{\right)}{\mathbf{=}}\frac{\mathbf{-}\mathbf{\u2206}{\mathbf{H}}_{\mathbf{rxn}}}{\mathbf{R}}\mathbf{[}\frac{\mathbf{1}}{{\mathbf{T}}_{\mathbf{2}}}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{T}}_{\mathbf{1}}}\mathbf{]}}$

We’re given the **ΔH˚ _{f} ** of each reactant and product:

**(CH _{3})_{2}CHOH (g)**

**(CH _{3})_{2}CO (g) **

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 ΔG_{rxn} at 570 K

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

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

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What scientific concept do you need to know in order to solve this problem?

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