## Problem 115 ch 11 6th edition

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

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

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

Raquel Ruiz 1K
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Joined: Tue Nov 14, 2017 3:02 am

### Problem 115 ch 11 6th edition

11.115 The overall photosynthesis reaction is 6 CO2(g) + 6 H2O(l) -> C6H12O6(aq) + 6 O2(g), and Delta H = +2802 kJ. Suppose that the reaction is at equilibrium. State the effect that each of the following changes will have on the equilibrium composition (tends to shift toward the formation of reactants, tends to shift toward the formation of products, or has no effect). (a) The partial pressure of O2 is increased. Can someone help me understand how to answer this question? Thank you !

Xuan Kuang 2L
Posts: 31
Joined: Wed Nov 14, 2018 12:23 am

### Re: Problem 115 ch 11 6th edition

In this case, we consider Le Chatelier's Principle. If the partial pressure of O2 were to increase, there would be more O2 molecules present, which results in the equilibrium ratio having a higher partial pressure of O2 than initially (Qp>Kp). To minimize this effect, the reaction will shift to the left, towards the formation of more reactants, to make Qp smaller.

(If it helps, you can view it in terms of concentration to make it conceptually easier :) )

jonathanjchang2E
Posts: 61
Joined: Fri Sep 28, 2018 12:26 am

### Re: Problem 115 ch 11 6th edition

The reaction shifts to the reactants side because more O2 is added and according to Le Chatelier's, adding a product should cause more reactants should be formed.