<Dd> − d (A) d t = k f (A) t − k b (B) t (\ displaystyle - (\ frac (d ((\ ce (A)))) (dt)) = (k_ (f) ((\ ce (A))) _ (t)) - (k_ (b) ((\ ce (B))) _ (t)) \,) </Dd> <P> The derivative is negative because this is the rate of the reaction going from A to B, and therefore the concentration of A is decreasing . To simplify annotation, let x be (A) t (\ displaystyle ((\ ce (A))) _ (t)), the concentration of A at time t . Let x e (\ displaystyle x_ (e)) be the concentration of A at equilibrium . Then: </P> <Dl> <Dd> − d (A) d t = k f (A) t − k b (B) t − d x d t = k f x − k b (B) t = k f x − k b ((A) 0 − x) = (k f + k b) x − k b (A) 0 (\ displaystyle (\ begin (aligned) - (\ frac (d ((\ ce (A)))) (dt)) & = (k_ (f) ((\ ce (A))) _ (t)) - (k_ (b) ((\ ce (B))) _ (t)) \ \ - (\ frac (dx) (dt)) & = (k_ (f) x) - (k_ (b) ((\ ce (B))) _ (t)) \ \ & = (k_ (f) x) - (k_ (b) ((\ ce ((A) 0)) - x)) \ \ & = ((k_ (f) + k_ (b)) x) - (k_ (b) (\ ce ((A) 0))) \ end (aligned))) </Dd> </Dl> <Dd> − d (A) d t = k f (A) t − k b (B) t − d x d t = k f x − k b (B) t = k f x − k b ((A) 0 − x) = (k f + k b) x − k b (A) 0 (\ displaystyle (\ begin (aligned) - (\ frac (d ((\ ce (A)))) (dt)) & = (k_ (f) ((\ ce (A))) _ (t)) - (k_ (b) ((\ ce (B))) _ (t)) \ \ - (\ frac (dx) (dt)) & = (k_ (f) x) - (k_ (b) ((\ ce (B))) _ (t)) \ \ & = (k_ (f) x) - (k_ (b) ((\ ce ((A) 0)) - x)) \ \ & = ((k_ (f) + k_ (b)) x) - (k_ (b) (\ ce ((A) 0))) \ end (aligned))) </Dd>

Expression for rate constant of second order reaction