Asample of nitrosyl bromide (nobr) decomposes according to the equation 2nobr(g)⇌2no(g)+br2(g)

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