Asked by luke
Consider the reaction at equilibrium:
2NO2(g) <---> N2O2(g) + O2(g)
for which, H = −20.4 kJ · mol−1. If you
decrease the temperature of the system, what will occur?
2NO2(g) <---> N2O2(g) + O2(g)
for which, H = −20.4 kJ · mol−1. If you
decrease the temperature of the system, what will occur?
Answers
Answered by
DrBob222
Rewrite the equations so the heat shows.
2NO2(g) <---> N2O2(g) + O2(g) + heat
Now, decreasing T means the reaction will try to heat the reaction. so it shifts to the ......
2NO2(g) <---> N2O2(g) + O2(g) + heat
Now, decreasing T means the reaction will try to heat the reaction. so it shifts to the ......
Answered by
luke
how do I know which side heat is on?
Answered by
DrBob222
In this case I wrote it on the correct side which is the right side. In any case however, remember that -delta H means the reaction is exothermic and + delta H means the reaction is endothermic. Look at this reaction; the problem tells you that delta H is -20 kJ/mol which means it is exothermic so you write heat as a product. If delta H is a + number it means the reaction is endothermic so you write heat as a reactant.
Answered by
please help
does the decrease mean the reaction shifts to the right?
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