Chemistry Thermodynamics Math Question 2

Question

What is the value of the equilibrium constant for transformation of carbon from graphite to diamond at 400.0 C?

SubstanceΔfH0	kJ/molS0	J/mol K
graphite	0.0	5.7
diamond	1.9	2.4

Answer 1

Table is messed up.

graphite ===> diamond is the reaction being considered at 400ºC = 673K

∆G = ∆H - T∆S

∆G for diamond = 1.9 kJ/mol - (673K x 0.0024 kJ/mol-K) = 1.9 - 1.6 = 0.3 kJ/mol

∆G for graphite = 0 - (673K x 0.0057 kJ/mol-K) = -3.8 kJ/mol

∆Grxn = ∆Gproducts - ∆Gractants = 0.3 - (-3.8) = 4.1 kJ/mol

∆G = -RT ln Keq

ln Keq = -∆G/RT

ln Keq = -4.1/(8.314)(673) = -0.000733

Keq = 0.999