Question

If a reaction is reversible, when can it be said to have reached equilibrium?

Answer 1

When the rate of the forward reaction is EQUIVALENT to the rate of the reverse reaction.............

Explanation:

For the reaction,

`A(g)+B(g)rightleftharpoonsC(g) + D(g)`

There is certainly a `"forward rate"=k_f[A][B]`

And a backwards rate, `"reverse rate"=k_r[C][D]`.

And by definition, `"chemical equilibrium"` specifies NOT the cessation of chemical change, BUT `"EQUALITY of FORWARD"` and `"REVERSE RATES."`

And thus at equilibrium, `k_f[A][B]=k_r[C][D]`, AND

`k_f/k_r=([C][D])/([A][B])`

And, clearly, if the forward rate is FASTER than the reverse rate, products are favoured at equilibrium..........and vice versa.

The quotient, `k_f/k_r` is better known as `K_"eq"`, the thermodynamic equilibrium constant, which is a constant according to temperature.

And so.........#k_f/k_r=K_"eq"=([C][D])/([A][B])#, and this is an equation with which you will get VERY familiar.........