A sample 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

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