A sample of nitrosyl bromide (NOBr) decomposes according to the equation 2NOBr(

Question

A sample of nitrosyl bromide (NOBr) decomposes according to the equation
2NOBr(g)2NO(g)+Br2(g)
An equilibrium mixture in a 5.00-L vessel at 100 C contains 3.29 g of NOBr, 3.07 g of NO, and 8.18 g of Br2.

Part A

Calculate Kc.

Part B

What is the total pressure exerted by the mixture of gases?

Express your answer to three significant figures and include the appropriate units.

Part C

What was the mass of the original sample of NOBr?

Express your answer to four significant figures and include the appropriate units.

Explanation / Answer

Let us find the number of moles of all compund at equilibrium

number of moles of NOBr = mass / molar mass
                                                       =3.29/110
                                                       = 0.03 mol
[NOBr] = number of moles/volume
               = 0.03/5
                =0.006 M

number of moles of NO= mass / molar mass
                                                       =3.07/30
                                                       = 0.102 mol
[NO] = number of moles/volume
               = 0.102/5
                =0.02 M

number of moles of Br2= mass / molar mass
                                                       =8.18/160
                                                       = 0.05 mol
[Br2] = number of moles/volume
               = 0.05/5
                =0.01 M

A)
Kc = [Br2][NO]^2 / [NOBr]^2
      = 0.01 * (0.02)^2/ (0.006)^2
      = 0.11

B)
Total number of moles = 0.03+ 0.102 + 0.05 = 0.18 mol

use:
P*V = n*R*T
P*5 = 0.18*(0.0821)*(100+273)
P= 1.1 atm

C)
2NOBr(g)<--->2NO(g)+Br2(g)
n                            0               0     (initial moles)
n-2x                      2x             x    (moles at eqiuilibrium)
clearly x = 0.05 mol
n-2x = 0.03
n - 2*0.05 = 0.03
n= 0.13 mol

molar mass of NoBr = 110 g/mol
Mass of NoBr = number of moles * molar mass
                             = 0.13 * 110
                             = 14.3 g

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