# For the equilibrium reaction X + y -> XY, the rate of the forward reaction at equilibrium is equal to 1.75 sec^(-1). What is the rate of the reverse reaction?

Nov 18, 2017

Based on the Law of Mass action, at Equilibrium the rate of the forward reaction equals the rate of the reverse reaction.

#### Explanation:

For the listed hypothetical reaction

Rate (fwd) = ${k}_{\text{fwd}} \left[X\right] \left[Y\right]$
Rate (rev) = ${k}_{\text{rev}} \left[X Y\right]$

By the 'Law of Mass Action' ...

Rate(fwd) = Rate(rev)

=> ${k}_{\text{fwd}} \left[X\right] \left[Y\right]$ = ${k}_{\text{rev}} \left[X Y\right]$

=> ${k}_{\text{fwd"/k_"rev}}$ = $\frac{\left[X Y\right]}{\left[X\right] \left[Y\right]}$ = Constant

=> The 'Constant' is defined as the Rxn Equilibrium Constant $\left({K}_{e q}\right)$ and is a relative measure of 'Extent of Reaction' or, how long it takes to move through the kinetic region of the concentration vs time trend to achieve constant concentrations of reactant and product. This is the 'Equilibrium Region' where Rate of Rxn fwd equals Rate of Rxn rev.