5.55

$\ln K = -\frac{\Delta H^{\circ}}{RT} + \frac{\Delta S^{\circ}}{R}$

Letty Liu 2C
Posts: 105
Joined: Fri Aug 09, 2019 12:16 am

5.55

C(s) 1 H2O(g) ∆CO(g) 1 H2(g). (a) Evaluate K At 900 K, given that the standard Gibbs free energies of formation of CO(g) and H2O(g) at 900 K are 2191.28 kJ?mol21 and 2198.08 kJ?mol21, respectively. (b) A sample of graphite of mass 5.20 kg and 125 g of water were placed into a 10.0-L container and heated to 900 K. What are the equilibrium concentrations?

Whenever we're not given the standard Gibbs free energy of molecules, do we assume it's zero?

KeyaV1C
Posts: 103
Joined: Sat Aug 17, 2019 12:17 am
Been upvoted: 1 time

Re: 5.55

All elements in their standard states will have 0 as their standard Gibbs free energy of formation.
Attachments

Aman Sankineni 2L
Posts: 103
Joined: Fri Aug 30, 2019 12:17 am

Re: 5.55

The standard Gibbs free energy of molecules is 0 for any element in its standard state.