Question

The reversible reaction, A(aq) + H2O(l) --> P(aq)

is first-order in A(aq) and P(aq), and pseudo-zero order in H2O(l). What must be

the most appropriate expression for the relaxation time, τ, when the reaction

returns to its equilibrium after a sudden increase in temperature?


the key answer is relaxation time=1/kf+kr but i don't know why the got rid of A with Kf and P with Kr

Answer 1

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 $\tau= \dfrac{1}{k_r+k_f}$