As water has an order of zero so neglect the water and the equation can be written as
"A\\iff" "P"
after equilibrium state is achieved and assuming the distance between the current state and the new equilibrium state is x , we get
"[A]=[A]_{eq}+x;[P]=[P]_{eq}-x;"
"d[A]\/dt=-k_r(x+[A]_{eq})+k_f([P]_{eq}-x)=-(k_r+k_f)x"
because "d[A]\/dt=dx\/dt" we get "dx\/dt" ="-(k_r+k_f)x"
therefore
"x=x_0e^{-t\/ \\tau}"
"1\/\\tau=k_r+k_f"
hence "\\tau= \\dfrac{1}{k_r+k_f}"
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