Chemistry

Chemistry

Problem

The reversible decomposition F2(g)⇌2 F(g)\mathrm{F_2(g)} \rightleftharpoons 2\,\mathrm{F(g)} is carried out in a sealed, rigid container. The standard free-energy change is ΔG∘=+4.8 kJ mol−1\Delta G^\circ = +4.8\,\text{kJ mol}^{-1} at 400 K400\,\text{K}. Assume ΔG∘\Delta G^\circ is temperature-independent over the range considered. Throughout this problem, work without a calculator. a. Part 1: At 400 K400\,\text{K}, estimate the equilibrium constant, K400K_{400}, to one significant figure and state whether the container will contain mostly F2\mathrm{F_2}, mostly F\mathrm{F}, or roughly equal amounts. Show your reasoning. b. Part 2: The chemist wishes to obtain at least 75%75\% atomic fluorine (F\mathrm{F}) at equilibrium. Determine---using approximate calculations---whether raising the temperature to 800 K800\,\text{K} will meet this goal. Clearly outline every assumption made and justify each estimation step. c. Part 3: Identify one experimental complication that could invalidate the assumptions made in part (b) and explain its potential effect on the predicted equilibrium composition.
Intuitive explanation

This problem asks you to predict which direction a reaction shifts and how far it goes, using the relationship between ΔG∘\Delta G^\circ, the reaction quotient QQ, and equilibrium. Since ΔG∘\Delta G^\circ is positive, the reaction doesn't favor products at standard conditions—but that doesn't mean no F atoms form. The key insight is that ΔG∘\Delta G^\circ tells you the thermodynamic tendency, while QQ tells you where you actually are; the system will shift to make ΔG\Delta G negative and reach equilibrium. You'll need to use ΔG∘=−RTln⁡K\Delta G^\circ = -RT \ln K to find the equilibrium constant, then reason about the extent of dissociation without heavy calculation.

Intuitive explanation

This problem asks you to predict which direction a reaction shifts and how far it goes, using the relationship between ΔG∘\Delta G^\circ, the reaction quotient QQ, and equilibrium. Since ΔG∘\Delta G^\circ is positive, the reaction doesn't favor products at standard conditions—but that doesn't mean no F atoms form. The key insight is that ΔG∘\Delta G^\circ tells you the thermodynamic tendency, while QQ tells you where you actually are; the system will shift to make ΔG\Delta G negative and reach equilibrium. You'll need to use ΔG∘=−RTln⁡K\Delta G^\circ = -RT \ln K to find the equilibrium constant, then reason about the extent of dissociation without heavy calculation.

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