Answer:
1) 17.6 x 10-3mol/L;
2) 857.3 Tor;
3)14.59 g
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
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
