Question

A sample of NOBr decomposes according to the following equation: 2 NOBr(g) 2 NO(g) + 1 Br2(g) An equilibrium mixture in a 6-L vessel at -173 oC, contains 0.493 g of NOBr, 0.140 g of NO, and 0.495 g of Br2.

(a) Calculate KP for this reaction at this temperature. KP = .

(b) What is the total pressure exerted by the equilibrium mixture of gases? Ptotal = atm.

Answer 1

0.493 g of NOBr = 0.493 g/109g /mol = 0.0045 moles of NOBr

0.140 g of NO = 0.140 g/30g/mol = 0.0046 moles of NO

0.495 g of Br₂ = 0.495 g / 158g/mol = 0.00313 moles of Br2

Now, convert the number of moles to concentration (moles per liter) dividing by 5 L :

[NOBr] = 0.0009 M

[NO] = 0.00092 M

[Br₂] = 0.000626 M

Reaction:

2NOBr(g) <---------> 2NO(g) + Br₂(g)

Kc = [Br₂] [NO]² / [NOBr]²

Kc =  0.000626 M * (0.00092 M)² / (0.0009 M)² = 6.54*10^-4

Kp =Kc *(RT)^delta n

Change in moles=3-2=1

Kp =6.54*10^-4 *0.082*100

=0.00536

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(b)

T = 100 K

Number of moles =0.002446 moles

V = 5 L

PV = nRT

P = nRT/V

P = 0.002446 x 0.082 x 100/5 = 0.004 atm