The equilibrium constant (Kc)for reaction at 80?C.If the

Chapter 14, Problem 4RC

(choose chapter or problem)

The equilibrium constant \(\left(K_{c}\right)\) for reaction \(A\ \leftrightharpoons\ B+C\) is \(4.8 \times 10^{-2}\) at \(80^{\circ} \mathrm{C}\). If the forward rate constant is \(3.2 \times 10^{2}\ \mathrm{s}^{-1}\), calculate the reverse rate constant.

Unfortunately, we don't have that question answered yet. But you can get it answered in just 5 hours by Logging in or Becoming a subscriber.

Becoming a subscriber
Or look for another answer

×

Login

Login or Sign up for access to all of our study tools and educational content!

Forgot password?
Register Now

×

Register

Sign up for access to all content on our site!

Or login if you already have an account

×

Reset password

If you have an active account we’ll send you an e-mail for password recovery

Or login if you have your password back