## 5G.17

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

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

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

Junwei Sun 4I
Posts: 125
Joined: Wed Oct 02, 2019 12:16 am

### 5G.17

The question asks to draw a graph that depicts the reaction in question 5G.13 which is I2(g) --> 2I(g) at 1200.K (K=6.8) when the partial pressure of I2 and I are 0.13 bar and 0.98 bar respectively. The answer key gives this graph but I still don't quite get what it means. Could someone please explain?
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Ryan Narisma 4G
Posts: 104
Joined: Fri Aug 30, 2019 12:18 am

### Re: 5G.17

Hi Junwei Sun 4I! To answer your question, because the reaction quotient value is greater than the value of K at this temperature, the reaction will proceed to form more reactants (I2 gas). The graph is depicting this instance. As time progresses, the reaction will start to form more I2, hence the increase in the partial pressure of I2. However, in doing so, the reaction must use up I (g) and this is depicted with the partial pressure of I (g) decreasing. Then when the reaction reaches equilibrium the partial pressures of each species remain unchanged, hence the leveling off of the partial pressures of each species after reaching a certain time. I hope this helps!

Junwei Sun 4I
Posts: 125
Joined: Wed Oct 02, 2019 12:16 am

### Re: 5G.17

Ryan Narisma 4G wrote:Hi Junwei Sun 4I! To answer your question, because the reaction quotient value is greater than the value of K at this temperature, the reaction will proceed to form more reactants (I2 gas). The graph is depicting this instance. As time progresses, the reaction will start to form more I2, hence the increase in the partial pressure of I2. However, in doing so, the reaction must use up I (g) and this is depicted with the partial pressure of I (g) decreasing. Then when the reaction reaches equilibrium the partial pressures of each species remain unchanged, hence the leveling off of the partial pressures of each species after reaching a certain time. I hope this helps!

Thank you!