Hi guys,
I am confused about the last question of this problem. Why does raising the temperature increase the k of the reaction with a greater Ea more than the rxn with the lower Ea? Can someone please explain?
Thanks
sapling #17
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Re: sapling #17
Since the reaction with the greater Ea has a greater energy hump to overcome than its reverse reaction, raising the temperature will increase its K more because it essentially decreases the value of that Ea by making the necessary leap in energy lower for that reaction. The way I like to think of it is that the reaction with the greater Ea is the endothermic reactions, and endothermic reactions need to consume energy from their surroundings. By increasing the temperature, you're giving them more ambient energy to consume (and by doing so you're decreasing the jump to the top of their activation energy hump). Thus, the reaction can happen more frequently and its K will increase.
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Re: sapling #17
Raising temperature lowers the activation energy which is the barrier that the reaction must overcome to occur. Because in the question, the forward rate's Ea is greater, more temperature input is needed to increase the rate, while for the reverse, it does not affect it as much, because it already has a lower activation energy. I would consider the reverse reaction to be exothermic, while the forward is endothermic, as it needs an input of heat.
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