Answer:
1) 17.6 x 10-3mol/L;
2) 857.3 Tor;
3)14.59 g
Step-by-step explanation:
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 3.27 g of NOBr, 3.09 g of NO, and 8.23 g of Br2.
1) Calculate Kc
.2) What is the total pressure exerted by the mixture of gases?
3) What was the mass of the original sample of NOBr
The above can be the concluding part to this question
1) Moles = Mass/Mol.mass
Moles NOBr=3.27 g/ 109.8g/mol = 0.0298 mol of NOBr
Moles NO=3.09 g/ 29.9g/mol = 0.1033mol NOMoles
Br2=8.23g/ 159.9g/mol = 0.0515mol Br2
Concentration of NOBr= 0.0298 mol/ 5.00 L = 0.00596 mol/L
Concentration of NO= 0.1033mol/ 5.00 L = 0.02066mol/L
Concentration of Br2= 0.0515mol/ 5.00 L = 0.0103mol/L
Kc for the equation = [NO]^2 * [Br2] / [NOBr]^2
Kc= [0.02066mol/L]^2 * [0.0103mol/L] / [0.00596 mol/L]^2=
17.6 x 10-3mol/L
From i deal gas equation of state
PV=nRT
P=nRT/V
find the total of the moles involved in the reaction
n=0.0298 mol+0.1033mol+ 0.0515mol= 0.1846 mol
T = 100 + 273= 373K(absolute temperature)
P= 0.1846 molx 373 K x 8.3 J/molK / 0.005 m3=
114300. 6 Pa=
857.3 Tor
3) conservation of mass m(NOBr)= 3.27 g+3.09 g +8.23 g = 14.59 g
1) 17.6 x 10-3mol/L;
2) 857.3 Tor;3)14.59g