Question #48db7

1 Answer
Jul 26, 2017

Answer:

#"Equality of forward and reverse rates of reaction......."#

Explanation:

For a generalized reaction............

#A+BrightleftharpoonsC+D#

There is a #"rate forward"=k_f[A][B]#....

And a #"rate backwards"=k_r[C][D]#....

#k_f# and #k_r# are simply rate constants that must be measured by experiment. The condition of chemical equilibrium DOES NOT specify cessation of chemical change BUT equality of forward and reverse rates........And thus.....

#"rate forward"=k_f[A][B]-="rate backwards"=k_r[C][D]#....

And so on rearrangement.......

#k_f/k_r="rate forwards"/"rate backwards"=([C][D])/([A][B])#

And the quotient #k_f/k_r# is otherwise known as the #"thermodynamic equilibrium constant"#, #K_"eq"#, which MUST be measured for a particular reaction.

And while, I admit, so far we have been very abstract we can clearly surmise that if #K_"eq"# is numerically LARGE, then the products #[C]# and #[D]# are favoured at equilibrium. And likewise, if #K_"eq"# is numerically SMALL, then the reactants #[A]# and #[B]# are favoured at equilibrium.

And if you can digest all that (and I don't know at which level you are) with understanding, then congratulations, you now understand #"chemical equilibrium"# to undergraduate level. Anyway, you have to see how to use this knowledge in the questions you are likely to be asked. And the definition you must know is that #"chemical equilibrium"# specifies equality of forward and reverse rates of reaction.