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I think the best way to think of equilibria intuitively is in terms of rates of reaction. At equilibrium, the forward and the reverse reactions are happening at the same rate.

If you increase the temperature, what happens to the rates of the forward and reverse reactions?

Using the Arrhenius equation

$$k = Ae^{-E_\mathrm{a}/(RT)}$$

you can see that as temperature increases, the rate will increase. The amount it increases depends on the activation energy ($E_\mathrm{a}$).

The activation energy on the endothermic side of the reaction will always be larger than the exothermic side. This is because the transition state is always higher in energy than either of the reactants/products, and the reactants for the endothermic reaction are by definition lower in energy than the products! This means that the endothermic reaction will be sped up more than the exothermic reaction. This means the reaction is now out of equilibrium and the endothermic reaction will happen until the reaction is back in equilibrium again.

I think the best way to think of equilibria intuitively is in terms of rates of reaction. At equilibrium, the forward and the reverse reactions are happening at the same rate.

If you increase the temperature, what happens to the rates of the forward and reverse reactions?

Using the Arrhenius equation, $$k = A\exp\left(\frac{-E_\mathrm{a}}{RT}\right),$$ you can see that as temperature increases, the rate will increase. The amount it increases depends on the activation energy $E_\mathrm{a}$.

The activation energy on the endothermic side of the reaction will always be larger than the exothermic side. This is because the transition state is always higher in energy than either of the reactants/products, and the reactants for the endothermic reaction are by definition lower in energy than the products! This means that the endothermic reaction will be sped up more than the exothermic reaction. This means the reaction is now out of equilibrium and the endothermic reaction will happen until the reaction is back in equilibrium again.

I think the best way to think of equilibria intuitively is in terms of rates of reaction. At equilibrium, the forward and the reverse reactions are happening at the same rate.

If you increase the temperature, what happens to the rates of the forward and reverse reactions?

Using the Arrhenius equation: you can see that as temperature increases, the rate will increase. The amount it increases depends on the Activation energy (Ea).

The activation energy on the endothermic side of the reaction will always be larger than the exothermic side. This is because the transition state is always higher in energy than either of the reactants/products, and the reactants for the endothermic reaction are by definition lower in energy than the products! This means that the endothermic reaction will be sped up more than the exothermic reaction. This means the reaction is now out of equilibrium and the endothermic reaction will happen until the reaction is back in equilibrium again.

I think the best way to think of equilibria intuitively is in terms of rates of reaction. At equilibrium, the forward and the reverse reactions are happening at the same rate.

If you increase the temperature, what happens to the rates of the forward and reverse reactions?

Using the Arrhenius equation

$$k = Ae^{-E_\mathrm{a}/(RT)}$$

you can see that as temperature increases, the rate will increase. The amount it increases depends on the activation energy ($E_\mathrm{a}$).

The activation energy on the endothermic side of the reaction will always be larger than the exothermic side. This is because the transition state is always higher in energy than either of the reactants/products, and the reactants for the endothermic reaction are by definition lower in energy than the products! This means that the endothermic reaction will be sped up more than the exothermic reaction. This means the reaction is now out of equilibrium and the endothermic reaction will happen until the reaction is back in equilibrium again.

I think the best way to think of equilibria intuitively is in terms of rates of reaction. At equilibrium, the forward and the reverse reactions are happening at the same rate.

If you increase the temperature, what happens to the rates of the forward and reverse reactions?

Using the Arrhenius equation: you can see that as temperate increases, the rate will increase. The amount it increases depends on the Activation energy (Ea).

The activation energy on the endothermic side of the reaction will always be larger than the exothermic side. This is because the transition state is always higher in energy than either of the reactants/products. This means that the endothermic reaction will be sped up more than the exothermic reaction. This means the reaction is now out of equilibrium and the endothermic reaction will happen until the reaction is back in equilibrium again.

I think the best way to think of equilibria intuitively is in terms of rates of reaction. At equilibrium, the forward and the reverse reactions are happening at the same rate.

If you increase the temperature, what happens to the rates of the forward and reverse reactions?

Using the Arrhenius equation: you can see that as temperature increases, the rate will increase. The amount it increases depends on the Activation energy (Ea).

The activation energy on the endothermic side of the reaction will always be larger than the exothermic side. This is because the transition state is always higher in energy than either of the reactants/products, and the reactants for the endothermic reaction are by definition lower in energy than the products! This means that the endothermic reaction will be sped up more than the exothermic reaction. This means the reaction is now out of equilibrium and the endothermic reaction will happen until the reaction is back in equilibrium again.