Nitrosyl bromide (NOBr) can be obtained as a pure liquid at low temperatures. Th

Question

Nitrosyl bromide (NOBr) can be obtained as a pure liquid at low temperatures. The liquid boils at -2 degree C, and at room temperature the gas partially decomposes, as shown below. A 2.00-g sample of cold liquid NOBr is injected into a 1.0 L flask. (Assume that the flask was first evacuated so that it does not contain any air or other gases.) When the flask is allowed to come to equilibrium at 298 K. the total pressure inside is measured as 0.624 atm. Calculate the total number of moles of gas present in the flask at equilibrium. (This part should be easy!) Now find the numerical value of K_eq for the reaction above at 298 K. SHOW ALL YOUR WORK IN AN ORGANIZED FASHION!

Explanation / Answer

a) we know that

PV = nRT

given

P = 0.624

V = 1

T = 298

so

0.624 x 1 = n x 0.0821 x 298

n = 0.0255

so

total number of moles of gas is 0.0255

b)

we know that

moles = mass/ molar mass

so

moles of NOBr = 2 / 110 = 0.01818

conc = moles / volume

[NObr]= 0.01818 / 1 = 0.01818

now the reaction is

2 N0Br ---> 2 N0 + Br2

now using ICE table

initial conc of NOBr , NO , Br2 are 0.01818 , 0 , 0

change in conc of NOBr , NO , Br2 are -2x , 2x , x

equilibrium conc of NOBr , NO , Br2 are 0.01818 -2x , 2x , x

as volume = 1 L

moles = concentration

so

total moles at equilibrium = 0.01818 -2x + 2x + x = 0.01818 + x

now

0.01818 + x = 0.0255

x = 0.00732

now

[NOBr] = 0.01818 - ( 2 * 0.00732) = 0.00354

[NO] = 2x = 2 * 0.00732 = 0.01464

[Br2] = x = 0.00732

now

Keq = [NO]^2 [ Br2] / [NOBr]^2

Keq = [ 0.01464]^2 [0.00732] / [ 0.00354]^2

Keq = 0.125

so

the value of Keq is 0.125

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