What Are the Equilibrium Concentrations and Kc for 2NO(g) + Br2(g) ⇌ 2NOBr(g)?

Click For Summary

Discussion Overview

The discussion revolves around calculating equilibrium concentrations and the equilibrium constant (Kc) for the reaction 2NO(g) + Br2(g) ⇌ 2NOBr(g). Participants explore the initial concentrations, changes in concentration, and the equilibrium state, as well as the correct expression for Kc.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Participants present initial concentrations for NO, Br2, and NOBr, and discuss the changes to find equilibrium concentrations.
  • Some participants propose values for P, Q, R, and S, with one stating P=0.5, Q=0.25, R=4.75, and S=1.5.
  • There is a question regarding the correct expression for Kc, with some suggesting Kc=[NOBr]^2 / ([NO]^2 [Br2]) and others questioning this formulation.
  • One participant provides a detailed calculation for equilibrium concentrations based on the extent of reaction, leading to a proposed Kc value of 2.671875.
  • Concerns are raised about the direction of the reaction arrow and its implications for the equilibrium expression.

Areas of Agreement / Disagreement

Participants express differing views on the correct equilibrium constant expression and the values for Q and R. The discussion remains unresolved regarding the correct formulation of Kc and the implications of the reaction direction.

Contextual Notes

Some assumptions regarding the extent of reaction and the definitions of concentrations may not be fully articulated, leading to potential discrepancies in calculations. The direction of the reaction and its effect on Kc is also a point of contention.

a7med2009
Messages
8
Reaction score
0
Given 2NO(g) + Br2(g) <--------_____> 2NOBr(g),

Concentration (M) [NO] [Br2] [NOBr]
Initial 2.5 5.0 1.0
Change P Q +0.5
Equilibrium 2.0 R S

Find P,Q,R,S and Give the correct equilibrium constant expression for the above reaction and
calculate Kc.
 
Physics news on Phys.org
a7med2009 said:
Given 2NO(g) + Br2(g) <--------_____> 2NOBr(g),

Concentration (M) [NO] [Br2] [NOBr]
Initial 2.5 5.0 1.0
Change P Q +0.5
Equilibrium 2.0 R S

Find P,Q,R,S and Give the correct equilibrium constant expression for the above reaction and
calculate Kc.

What is the difference between 2.5 and 2.0?
 
sjb-2812 said:
What is the difference between 2.5 and 2.0?

I got the first part
p=0.5 ,Q=0.25, R=4.75 , s=1.5

but for Kc
is it Kc=[NOBr]^2 /[NO]^2 [Br2]

or Kc=[NO]^2 [Br2]/[NOBr]^2
 
What is the generic equation for any equilibrium constant, say A + B <-> C + D?

Are you sure about Q & R?
 
sjb-2812 said:
What is the generic equation for any equilibrium constant, say A + B <-> C + D?

Are you sure about Q & R?

2NO(g) + Br2(g) <----___> http://www.freeimagehosting.net/uploads/b72c8010cb.jpg 2NOBr(g),
2.5 5.0 1.0
Since the total change in conc of [NOBr] is given as +0.5
so x[ extant of reaction]=0.50M (2mole NO to 2mole NOBr)
Now using the general expression
At eqm [NO] =intial conc- extant of reaction =2.5-0.5=2.0M
[Br2] = initial conc-(extant of reaction/2) =5.0-(0.5/2)=4.75M
[NOBr] = initial conc+ extant of reaction =1.0+0.5 =1.5M
So P=0.5M Q= 0.25 M R= 4.75M S=1.5M
Kc =[NOBr]2/[NO]2*[Br2]= (1.5)2/(2.0)2*(4.75) =2.671875

(b) Give the correct equilibrium constant expression for the above reaction and
calculateKc.
Kc =[NOBr]2/[NO]2*[Br2]= (1.5)2 (morality)2/((2.0)2(morality)2 *(4.75) (molarity))=2.671875 molar-1
Unit of Kc= liter.mole-1

is it correct , the problem is that the arrow is going from right first the coming back (from
2NOBr(g) first to 2NO(g) + Br2(g) ) I'm confused,,,
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
Replies
24
Views
4K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
2
Views
10K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 11 ·
Replies
11
Views
5K