11.15 Gibbs Free Energy and Q

$\Delta G^{\circ}= \Delta H^{\circ} - T \Delta S^{\circ}$

$\Delta G^{\circ}= -RT\ln K$

$\Delta G^{\circ}= \sum \Delta G_{f}^{\circ}(products) - \sum \Delta G_{f}^{\circ}(reactants)$

April P 1C
Posts: 38
Joined: Thu Jan 12, 2017 3:01 am

11.15 Gibbs Free Energy and Q

11.15 (a) Calculate the reaction Gibbs free energy of I2(g) -> 2 I(g) at 1200. K (K=6.8) when the partial pressures of I2 and I are 0.13 bar and 0.98 bar, respectively.

I used deltaG* =-RTlnK
and then used
deltaG=deltaG*+RTlnQ

I'm not getting the correct answer though. How do I calculate Q?

Yashaswi Dis 1K
Posts: 56
Joined: Fri Sep 29, 2017 7:04 am

Re: 11.15 Gibbs Free Energy and Q

Hi,

Yeah so to find Q, you can just plug in the partial pressures which is directly proportional to concentrations, so Q = (.98)^2/(.13). I hope that helps! That should then give you the right answer hopefully.

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