[XGen]

A reaction has a forward rate constant of 2.3 x 10⁶ and an equilibrium constant of 4.0 x 10⁸. What is the rate constant for the reverse reaction?

Letting A, C, D, and E be arbitrary reactants and products, I have k_f[A][C] = k_b[D][E]. Solving for k_b, I get k_b = k_f([A][C]/[D][E]). This can also be simplified into k_b = k_f(1/K_eq), because K_eq is equal to ([D][E]/[A][C]). Plugging in the values, I get k_b = (2.3 x 10⁶)(1/4.0 x 10⁸), or 5.8 x 10^-3. However, the answer key says this is wrong.

Can anyone point out where I went wrong?

[sjb]

A reaction has a forward rate constant of 2.3 x 10⁶ and an equilibrium constant of 4.0 x 10⁸. What is the rate constant for the reverse reaction?

Letting A, C, D, and E be arbitrary reactants and products, I have k_f[A][C] = k_b[D][E]. Solving for k_b, I get k_b = k_f([A][C]/[D][E]). This can also be simplified into k_b = k_f(1/K_eq), because K_eq is equal to ([D][E]/[A][C]). Plugging in the values, I get k_b = (2.3 x 10⁶)(1/4.0 x 10⁸), or 5.8 x 10^-3. However, the answer key says this is wrong.

Can anyone point out where I went wrong?

I don't think you need the equilibrium constant, so that k_b k_f = 1

[Olympiad_Tutor]

A reaction has a forward rate constant of 2.3 x 10⁶ and an equilibrium constant of 4.0 x 10⁸. What is the rate constant for the reverse reaction?

Letting A, C, D, and E be arbitrary reactants and products, I have k_f[A][C] = k_b[D][E]. Solving for k_b, I get k_b = k_f([A][C]/[D][E]). This can also be simplified into k_b = k_f(1/K_eq), because K_eq is equal to ([D][E]/[A][C]). Plugging in the values, I get k_b = (2.3 x 10⁶)(1/4.0 x 10⁸), or 5.8 x 10^-3. However, the answer key says this is wrong.

Can anyone point out where I went wrong?

the answer key is wrong

unless additional details are given about the reaction

[qw098]

I believe the forward rate constant divided by the backward rate constant gives you the equilibrium constant so it is easy for you to solve for the equilibrium constant

[XGen]

I believe the forward rate constant divided by the backward rate constant gives you the equilibrium constant so it is easy for you to solve for the equilibrium constant

Your method gives my answer, which is incorrect on the answer key.

Thank you for your input, Ch_Olympiad_Winner. There is no other information given.

[Olympiad_Tutor]

sure.
when everything else fails we must accept the possibility of a typo.

as qw098 said the problem is easy.

if "can't be determined from the info given" is not the answer then it is definitely a typo.

[PIQgoogleme]

I like this problem. Your solution looks good. Where'd you find it? any more like it?

[XGen]

This is from the 1999 (or 2000? 2001? I forgot D:) USA National Chemistry Exam.

[PIQgoogleme]

Oh cool. I'm studying for that also. Good luck

[juanrga]

A reaction has a forward rate constant of 2.3 x 10⁶ and an equilibrium constant of 4.0 x 10⁸. What is the rate constant for the reverse reaction?

Letting A, C, D, and E be arbitrary reactants and products, I have k_f[A][C] = k_b[D][E]. Solving for k_b, I get k_b = k_f([A][C]/[D][E]). This can also be simplified into k_b = k_f(1/K_eq), because K_eq is equal to ([D][E]/[A][C]). Plugging in the values, I get k_b = (2.3 x 10⁶)(1/4.0 x 10⁸), or 5.8 x 10^-3. However, the answer key says this is wrong.

Can anyone point out where I went wrong?

Could you give more info? units? answer key?...