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

[a7med2009]

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.

 

[sjb-2812]

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?

 

[a7med2009]

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

 

[sjb-2812]

What is the generic equation for any equilibrium constant, say A + B <-> C + D?

Are you sure about Q & R?

 

[a7med2009]

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,,,