?At 700 K, the equilibrium constant for the reaction\(\mathrm{CCl}_{4}(g)

Chapter 15, Problem 15.89

(choose chapter or problem)

At 700 K, the equilibrium constant for the reaction

\(\mathrm{CCl}_{4}(g) \rightleftharpoons \mathrm{C}(s)+2 \mathrm{Cl}_{2}(g)\)

is \(K_{p}=0.76\). A flask is charged with 2.00 atm of \(\mathrm{CCl}_{4}\), which then reaches equilibrium at 700 K.

(a) What fraction of the \(\mathrm{CCl}_{4}\) is converted into C and \(\mathrm{Cl}_{2}\) ?

(b) What are the partial pressures of \(\mathrm{CCl}_{4}\) and \(\mathrm{Cl}_{2}\) at equilibrium?

Text Transcription:

CCl_4(g) \rightleftharpoons C(s)+2 Cl_2(g)

K_p=0.76

CCl_4

Cl_2

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