Consider the dissociation of methane to graphite (C) and hydrogen gas. For the methane at 208K, the data available are: ∆fH°CH4 = -74.85 kJ/ mol ∆fS°CH4 = -80.67 J/ mol K ∆fG°c,graphite = ∆fG° H2(g)=0 a. Calculate the equilibrium constant (K) of the dissociation of methane at 298 K with the appropriate units. b. Assuming that ∆H is independent of temperature, calculate the equilibrium constant (K) of methane dissociation at 1000 K c. Calculate the ratio between the amounts of methane and hydrogen gases at 298 K and 0.2 barg
Consider the dissociation of methane to graphite (C) and hydrogen gas. For the methane at 208K, the data available are: ∆fH°CH4 = -74.85 kJ/ mol ∆fS°CH4 = -80.67 J/ mol K ∆fG°c,graphite = ∆fG° H2(g)=0 a. Calculate the equilibrium constant (K) of the dissociation of methane at 298 K with the appropriate units. b. Assuming that ∆H is independent of temperature, calculate the equilibrium constant (K) of methane dissociation at 1000 K c. Calculate the ratio between the amounts of methane and hydrogen gases at 298 K and 0.2 barg
Chemistry & Chemical Reactivity
10th Edition
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Chapter18: Principles Of Chemical Reactivity: Entropy And Free Energy
Section18.7: The Interplay Of Kinetics And Thermodynamics
Problem 2.1ACP: The decomposition of diamond to graphite [C(diamond) C(graphite)] is thermodynamically favored, but...
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Consider the dissociation of methane to graphite (C) and hydrogen gas. For the methane at 208K, the data available are:
∆fH°CH4 = -74.85 kJ/ mol
∆fS°CH4 = -80.67 J/ mol K
∆fG°c,graphite = ∆fG° H2(g)=0
a. Calculate the equilibrium constant (K) of the dissociation of methane at 298 K with the appropriate units.
b. Assuming that ∆H is independent of temperature, calculate the equilibrium constant (K) of methane dissociation at 1000 K
c. Calculate the ratio between the amounts of methane and hydrogen gases at 298 K and 0.2 barg.
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