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1. See Question 4 (attached).
The answer is A, but why is it not B or D?
What's the rule here or the method to work this out?

Initially, I thought of using Surr = -H/T and that an increase in temperature decreases Surr entropy - but I don't know how to use this to help answer this question.
It says that a higher temperature moves the equilibrium to the side of products (RHS), so the reaction must be exothermic?

Help appreciated - thank you
Attached Files
2. Chapter 2.pdf (89.4 KB, 119 views)
3. (Original post by lekha2611)
See Question 4 (attached).
The answer is A, but why is it not B or D?
What's the rule here or the method to work this out?

Initially, I thought of using Surr = -H/T and that an increase in temperature decreases Surr entropy - but I don't know how to use this to help answer this question.
It says that a higher temperature moves the equilibrium to the side of products (RHS), so the reaction must be exothermic?

Help appreciated - thank you
If you increase the temperature of a system and the equilibrium shifts towards the products, the forward reaction is endothermic, as per Le Chatelier's Principle of opposing changes.

If a reaction is endothermic, delta H is positive. This rules out B and D.

According to the Gibbs Free Energy equation, delta G = delta H - (T delta S). In order for a reaction to be spontaneous, delta G must be negative. Therefore, (T delta S) must be greater than (delta H). Since (delta H) in this case is positive, (T delta S) must be even more positive. This means that (delta S) is positive. This rules out C.

4. (Original post by thegodofgod)
If you increase the temperature of a system and the equilibrium shifts towards the products, the forward reaction is endothermic, as per Le Chatelier's Principle of opposing changes.

If a reaction is endothermic, delta H is positive. This rules out B and D.

According to the Gibbs Free Energy equation, delta G = delta H - (T delta S). In order for a reaction to be spontaneous, delta G must be negative. Therefore, (T delta S) must be greater than (delta H). Since (delta H) in this case is positive, (T delta S) must be even more positive. This means that (delta S) is positive. This rules out C.

I thought that if you increase the temperature, the system wants to decrease the temperature and does so by move the equilibrium to the exothermic direction (endothermic would increase the temperature of the system, and not decrease it). So if you make the forward reaction exothermic, increasing the temperature will make more products.

But the Gibbs energy idea makes sense - thanks
5. (Original post by lekha2611)
I thought that if you increase the temperature, the system wants to decrease the temperature and does so by move the equilibrium to the exothermic direction (endothermic would increase the temperature of the system, and not decrease it). So if you make the forward reaction exothermic, increasing the temperature will make more products.

But the Gibbs energy idea makes sense - thanks
No, remember that if you move an equilibrium towards the side of the product and the forward reaction is endothermic, the temperature of the reaction vessel will decrease.

Do you remember the Haber Process? The forward reaction is exothermic (forming ammonia), which is why in industrial conditions the temperature is kept (relatively) low, in order to shift the equilibrium towards the side of the product, as this will oppose the reduction in temperature by increasing it with the formation of ammonia. However, if the temperature is kept too low, the rate of reaction will decrease, so the product yield will be low. Hence, the normal temperature is kept at around 450oc.
6. (Original post by thegodofgod)
No, remember that if you move an equilibrium towards the side of the product and the forward reaction is endothermic, the temperature of the reaction vessel will decrease.

Do you remember the Haber Process? The forward reaction is exothermic (forming ammonia), which is why in industrial conditions the temperature is kept (relatively) low, in order to shift the equilibrium towards the side of the product, as this will oppose the reduction in temperature by increasing it with the formation of ammonia. However, if the temperature is kept too low, the rate of reaction will decrease, so the product yield will be low. Hence, the normal temperature is kept at around 450oc.
I think I'm getting confused with the effect of temperature on the equilbrium.
If the forward reaction is exothermic, it's releasing heat to the surroundings not the system? So, if you increase the heat supplied to the system, it wants to cool the system, so it expells it to the surroundings i.e. using the forward reaction?

But you're saying the forward reaction is exothermic and so the temperature is kept low to form more products. If the temperature of the system is decreased, it wants to increase the temperature - so it does this by the forward reaction (increasing products) - so does this increase the temperature of the system or the surroundings - or both?
7. (Original post by lekha2611)
I think I'm getting confused with the effect of temperature on the equilbrium.
If the forward reaction is exothermic, it's releasing heat to the surroundings not the system? So, if you increase the heat supplied to the system, it wants to cool the system, so it expells it to the surroundings i.e. using the forward reaction?

But you're saying the forward reaction is exothermic and so the temperature is kept low to form more products. If the temperature of the system is decreased, it wants to increase the temperature - so it does this by the forward reaction (increasing products) - so does this increase the temperature of the system or the surroundings - or both?
No, you're getting confused

If the forward reaction is exothermic, the products produce heat and the reactants (when they act as products of the backward reaction) reduce the heat.

Therefore, if you increase the temperature, the equilibrium will shift to the left by producing more reactants, which would reduce the temperature.
8. (Original post by thegodofgod)
No, you're getting confused

If the forward reaction is exothermic, the products produce heat and the reactants (when they act as products of the backward reaction) reduce the heat.

Therefore, if you increase the temperature, the equilibrium will shift to the left by producing more reactants, which would reduce the temperature.
ahh okay, I think I get it now!
If the forward is exothermic, making products releases heat and so forming the reactants again will require heat, so it takes it in, thus reducing the temperature i.e. the backward reaction is endothermic?

So I guess if something is exothermic, it's releases heat from the system to the surroundings and the temperature is measured in the surroundings.
So, if the reaction is endothermic, the heat energy moves from the surroundings to the system, thus a temperature (measured in the surroundings) decreases?
9. (Original post by lekha2611)
ahh okay, I think I get it now!
If the forward is exothermic, making products releases heat and so forming the reactants again will require heat, so it takes it in, thus reducing the temperature i.e. the backward reaction is endothermic?

So I guess if something is exothermic, it's releases heat from the system to the surroundings and the temperature is measured in the surroundings.
So, if the reaction is endothermic, the heat energy moves from the surroundings to the system, thus a temperature (measured in the surroundings) decreases?

Perfect!
10. (Original post by thegodofgod)

Perfect!
Woo hoo! Thank you for your help

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