# EK Exam 1H Questions

Discussion in 'MCAT Study Question Q&A' started by JiniInBottle, Jul 28, 2011.

1. ### JiniInBottle 5+ Year Member

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for question 23:
CO + 1/2 O2 --> CO2 H= -283
At very high temp, rxn 1 will most likely:
I understand why it is nonspontaneous. But wouldn't it go in reverse direction if the temp is high b/c enthalpy is negative so it is like adding temp to the products so...but the answer doesn't consider that. It just looks at forward reaction, which I understand. But normally, if I'm right the reaction would go in reverse correct? I guess b/c this equation doesn't have the double arrow, I should have not though about the reverse reaction but only the forward even with high temp it would go forward only!!!1

2. OP

### JiniInBottle 5+ Year Member

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Also, do you think that EK exam is only a bit harder...than real?

3. ### BeatMCAT

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umm I would assume the same. As heat is added into the system, it should be absorbed, and that would be endothermix rxn, which would lead to reactants. Must be a mistake.

4. ### MShopes 5+ Year Member

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Heat is exothermic in the forward reaction and endothermic in the reverse reaction. This can be proven by the activation energy. Since the reaction is exothermic in the forward reaction, that means the activation energy of the forward reaction is less than that of the reverse reaction and heat is released from the system in the forward reaction but absorbed in the reverse reaction. At that moment, the reaction is leaning toward the right more and thus spontaneous at that moment. As more heat is added, the reaction starts shifting to the left and thus becomes nonspontaneous in the forward reaction or spontaneous in the reverse reaction.

Notice the reverse of forward and reverse reaction. If forward is exo, then reverse reaction is endo. If reverse is spontaneous, forward is nonspontaneous.

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