Nitrosyl bromide (NOBr) can be obtained as a pure liquid at low temperatures. Th
ID: 931287 • Letter: N
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
Explanation / Answer
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
[NOBr]
[NO]
[Br2]
initial
0.0182
0
0
change
-x
+x
+x
equilibrium
0.0182-x
x
x
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