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According to the article Summary of Le Chatelier's Principle on dynamicscience.com.au, for the process

$$\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)}\quad ΔH = \pu{-92 kJ/mol}$$

the forward reaction is exothermic, while the reverse reaction is endothermic.

If the forward reaction is exothermic, increasing temperature decreases $K,$ because it drives the equilibrium backwards, in order to absorb the heat energy.

However, in the graph, there is an abrupt change in the reaction rate of forward and reverse reactions. Why does the forward (exothermic) reaction rate rise up so abruptly, while the reverse (endothermic) reaction does not go up so high?

Shouldn't it clearly be the opposite, as increasing temperature favors the reverse reaction?

graph

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  • $\begingroup$ Dont try to learn chemistry off dodgy websites ;) $\endgroup$ – Karl Nov 26 '19 at 21:14
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Shouldn't it clearly be the opposite, as increasing temperature favors the reverse reaction?

Yes, they made a mistake. For all the other scenarios, they paired a figure of the rate changes with a figure of the matching concentration changes. For this scenario (increase in temperature) they matched it with a correct figure of concentration changes when the temperature is decreased:

enter image description here

If you change the label in the figure posted by the OP, it would reflect what happens when increasing the temperature: Both rates increase, but the reverse rate increases by a larger factor, resulting in a net reverse reaction until equilibrium is reached again.

Here are sketches for the rate changes both for increase in temperature (low to high, upper panel) and decrease in temperature (back to low temperature, lower panel). As you can see, the reverse rate is more sensitive to temperature (higher activation energy because the product is lower in energy than the reactant for this exothermic reaction):

enter image description here

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