Part 4 Post-Module Question

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Ashley Bouma 1F
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Joined: Fri Sep 28, 2018 12:21 am

Part 4 Post-Module Question

Postby Ashley Bouma 1F » Tue Jan 15, 2019 1:55 pm

I am confused on question 17 in the part 4 post module, could anyone please explain?
Determine the shift in equilibrium, if any, which will occur for the following reactions when the temperature is increased
a) Photosynthesis:
6 CO2 (g) + 6 H2O (l) ⇌ C6H12O6(s) + 6 O2 (g) delta H° = +2802 kJ.mol-1
b) The hydrolysis of ATP:
ATP (aq) + H2O (l) ⇌ ADP + PO42-(aq) delta H° = -30 kJ.mol-1

Tam To 1B
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Joined: Fri Sep 28, 2018 12:25 am

Re: Part 4 Post-Module Question

Postby Tam To 1B » Tue Jan 15, 2019 2:00 pm

In this problem, we are taking into account of exothermic and endothermic reactions. The reaction shifts toward the endothermic reaction.
If a reaction is endothermic, it will have a positive delta H, meaning that increasing temperature will favor the forward reaction and product formation.
If a reaction is exothermic, it will have a negative delta H, meaning that increasing temperature will favor the reverse reaction and reactant formation.

Fiona Jackson 1D
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Joined: Fri Sep 28, 2018 12:22 am

Re: Part 4 Post-Module Question

Postby Fiona Jackson 1D » Tue Jan 15, 2019 2:00 pm

Hi! The first equation is endothermic, since delta H is positive, so treat temperature as a reactant. If you increase temperature, or increase the amount of reactant, the reaction will shift right towards the products to restore equilibrium. The same logic can be used for the second equation, but since delta H is negative, it is exothermic and temperature should be treated as a product. I hope that helps!

A De Castro 14B 2H
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Joined: Fri Sep 28, 2018 12:29 am

Re: Part 4 Post-Module Question

Postby A De Castro 14B 2H » Tue Jan 15, 2019 2:01 pm

The delta H values indicate whether the reaction is endothermic or exothermic. When delta H is positive, the reaction is endothermic, simply meaning that heat is in the reactants side. When delta H is negative, the reaction is exothermic, simply meaning that heat is in the products side.

Thus, in a) when temperature is increased, heat is added, thus there are more reactants. According to Le Chatlier's principle, the equilibrium will shift to the right/products side. As for b), there will be more products when temperature is increased, so the equilibrium will shift to the left/reactants side.

805087225
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Joined: Thu Jun 07, 2018 3:00 am

Re: Part 4 Post-Module Question

Postby 805087225 » Tue Jan 15, 2019 6:31 pm

If delta H is positive, the increase in temperature will favour forward reaction, and if delta H is negative, it favours reverse reaction.


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