If we begin with equal of reactants and products, the forward reaction will be faster than the reverse because e^-3.61/RT is greater than e^-85.7/RT. More products will accumulate.

equilibrium, defined by rate_f = rate_r, will not be reached until the excess of products is great enough to compensate for the larger forward rate constant, which is a consequence of the smalleractivation energy of the forward reaction.

The equilibrium constant, which is the ratio of products to reactants at equilibrium, therefore will be larger than 1.00. If the products are thermodynamically less stable than reactants (positive DH⁰ of reaction), then the reverse rate constant will be greater than the forward, equilibrium will be attained with an excess of reactants, and K_eq will be less than 1.00. Both situations are diagrammed opposite.

It is always possible to increase a rate constant and accelerate a reaction by increasing the temperature. For reactions with an activation energy of 12 to 13 kcal, at temperatures around 298⁰K the rate constant doubles with every 10⁰ rise in temperature. (Can you prove this?)

But as we have seen with NH₃, there can be difficulties: The reverse reaction may be accelerated faster than the forward reaction, so that fewer products are obtained. The products or reactants may be unstable at elevated temperatures, or in special applications the surroundings may preclude the use of higher temperatures.