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    The platinum-rhodium catalyst is used to make NO, but without it, NH3 produces N2- why is NO favoured over N2?

    Why are exothermic reactions favoured in lower temperatures?

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    Exothermic reactions are favoured in lower temperatures because the system is trying to maintain normal conditions, it's just something you will need to learn; lower temperature, the system will shift in equilibrium quantities so that the temperature increases again. The same with pressure. If 2 moles of gas are the products and 1 mole the reactants, and the pressure is lowered, the system wants the pressure to rise again and so it will make more products, since the pressure of the products is higher than that of the reactants.

    Think of the "system" as the reaction and everything in in. I personify it to make it easier to understand, but you have to remember that there's no thought behind it - the real reason it happens it to be at the lowest energy state that it can.
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    Thanks for your help mik1a! I have a few Qs about what you said:

    When you say that at lower temperatures, the equilibrium will shift so that the temperature increases again..how does it increase again?

    Also, how would you tell if a reaction is exothermic (is it by the way the equilibrium shifts?).
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    (Original post by wigman15)
    Thanks for your help mik1a! I have a few Qs about what you said:

    1) When you say that at lower temperatures, the equilibrium will shift so that the temperature increases again..how does it increase again?

    2) Also, how would you tell if a reaction is exothermic (is it by the way the equilibrium shifts?).
    1) By making more of a product: the reaction is exothermic so by making more products more thermal energy is released from the "system" - all the compounds elements and molecules and their bonds.

    2) Do it, and if it get's hot it's exothermic
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    but how would you tell if you didn't do the experiment?
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    Well from never having done an experiment all you could do is predict, but we have information books that tell us enthalpies of bond formations (such as C-C = 393 or something kJ mol^-1), and from the equation of the reaction we can use the information to find the enthalpy of the current bond structure, and find the enthalpy of the final bond structure, subtract the first from the second and obtain the enthalpy change of the reaction.

    If this change is less than 0, the system has lost energy, and due to the law of conservation of energy, that energy goes somewhere - usually in the form of heat, therefore it is exothermic. The same if the enthalpy change is above 0, the system has gained energy from its surroundings and so it feels cold.
 
 
 
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