## 11.115

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

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

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

Wilson Yeh 1L
Posts: 42
Joined: Fri Sep 29, 2017 7:06 am

### 11.115

The overall photosynthesis reaction is 6 CO2(g) + 6 H2O(l) --> C6H12O6(aq) + 6 O2(g). Standard enthalpy of reaction is +2802 kJ. State the equilibrium composition.

The answer states for part B and part F that there is no effect when the system is compressed and water is added, respectively.

Why is that?

For part B, my initial answer was that the equilibrium composition would shift toward the formation of product because there's less overall moles. Is the answer no effect because only gas is taken into account when compressing and expanding so since there's the same number of moles of gas on each side then there's no effect?

For part F, I think the reasoning behind the fact that there is no effect on equilibrium composition when water is added is because there's water on the reactant side and because C6H12O6(aq) is an aqueous solution there's water on the product side also, so adding water doesn't affect the reaction.

I'm not sure if my logic is correct, can someone clarify for me? Thanks!

Curtis Wong 2D
Posts: 62
Joined: Sat Jul 22, 2017 3:00 am

### Re: 11.115

Basically yes. So this goes back to about our ideas of equilibrium constants. When you compress a gas (decrease the volume), the reaction will shift to the side with less moles of gas, but since there's the same amount of moles of gas on either side, there's no effect. For the water, we only take into account for our equilibrium constant gas and aqueous solutions, so adding solids or liquids would not affect the equilibrium constant and its shift towards one side or another.