11.43

Moderators: Chem_Mod, Chem_Admin

Jade Fosburgh Discussion 2C
Posts: 59
Joined: Fri Sep 29, 2017 7:07 am

11.43

Postby Jade Fosburgh Discussion 2C » Wed Nov 22, 2017 10:54 pm

Consider the reaction 2 NO(g) <--> N2(g) O2(g).

If the initial partial pressure of NO(g) is 1.0 bar, and x is the equilibrium concentration of N2(g), what is the correct equilibrium relation?

(a) K = x2/(1.0 - x)

(b) K = x2

(c) K = x2/(1.0 - 2x)2

(d) K = 4x3/(1.0 - 2x)2

(e) K = 2x/(1.0 - x)2.

I thought the answer would be (b). Can someone explain why you need to have 1.0 - 2x in the denominator?

Golbarg Rahimi 3k
Posts: 26
Joined: Sat Jul 22, 2017 3:01 am

Re: 11.43

Postby Golbarg Rahimi 3k » Thu Nov 23, 2017 3:41 pm

for this question, you have to draw the ICE table. the initial pressure of the reactant is 1. Its stoichiometry coefficient is 2 and if we consider the change of the products as X, the change NO undergoes would be -2X. so the pressure at equilibrium for NO would be (1-2x) and X for each of the products. Now the equilibrium equation would be k=x^2/ (1-2x)^2


Return to “Equilibrium Constants & Calculating Concentrations”

Who is online

Users browsing this forum: No registered users and 3 guests