Question

Nitrosyl bromide (NOBr) can be obtained as a pure liquid at low temperatures. The liquid boils at –2°C, and at room temperature the gas partially decomposes, as shown below.
2 NOBr = 2 NO(g) + Br2(g)
A 2.00-g sample of cold liquid NOBr is injected into a 1.0 L flask. When the flask is allowed to come to equilibrium at 298 K, the total pressure inside is measured as 0.624 atm
*There are .0255 moles of gas present in the flask at equilibrium

Keq for the reaction above at 298 K? (HINT: Set up the usual equilibrium table, and then try to relate the final concentrations to the total number of moles you found

Answer 1

2 NOBr = 2 NO(g) + Br2(g)

Moles of NOBr=2.00g/molar mass of NOBr=2.00g/109.9 g/mol=0.0182 moles

[NOBr=109.9 g/mol]

ICE table

	[NOBr]	[NO]	[Br2]
initial	0.0182	0	0
change	-x	+x	+x
equilibrium	0.0182-x	x	x

At equilibrium,

Total moles=0.0255 moles

0.0182-X+X+X=0.0255

Or,0.0182+X=0.0255

Or,X=0.0255-0.0182=0.0073 moles

[NOBr]=0.0182-x=0.0182-0.0073=0.0109 moles

[Br2]=[NO]=x=0.0073 moles

Keq=[NO]^2 [Br2]/[NOBr]^2=(0.0073 mol/1L)^2 *(0.0073 mol/1L)/(0.0109 mol/1L)^2

Keq=(0.0073 M)^3/(0.0109M)^2=0.003274=3.27*10^-3

Keq=3.27*10^-3M