Answer to Question #103415 in Physical Chemistry for lisa

Question #103415
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
1
Expert's answer
2020-02-21T07:52:29-0500

As water has an order of zero so neglect the water and the equation can be written as

A    A\iff PP

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]eqx;[A]=[A]_{eq}+x;[P]=[P]_{eq}-x;

d[A]/dt=kr(x+[A]eq)+kf([P]eqx)=(kr+kf)xd[A]/dt=-k_r(x+[A]_{eq})+k_f([P]_{eq}-x)=-(k_r+k_f)x

because d[A]/dt=dx/dtd[A]/dt=dx/dt we get dx/dtdx/dt =(kr+kf)x-(k_r+k_f)x

therefore

x=x0et/τx=x_0e^{-t/ \tau}

1/τ=kr+kf1/\tau=k_r+k_f

hence τ=1kr+kf\tau= \dfrac{1}{k_r+k_f}


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment