Asample 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 3.29 g of nobr, 3.04 g of no, and 8.10 g of br2. part apart complete calculate kc. express your answer using three significant figures. 0.116 previous answers correct part bpart complete what is the total pressure exerted by the mixture of gases? express your answer to three significant figures and include the appropriate units. 1.11 atm previous answers correct part c what was the mass of the original sample of nobr?

a. Kc = 0,116

b. 1,11 atm

c. 25,5 g of NOBr

Explanation:

a. For the equilibrium:

2NOBr(g)⇌2NO(g)+Br₂(g)

kc is defined as:

(1)

The molar concentration of each compound are:

NOBr: 3,29 g× = 0,0299 moles/ 5,00L = 5,986x10⁻³M

[NO]: 3,04 g× = 0,101 moles/ 5,00L = 0,0203M

[Br₂]: 8,10 g× = 0,0507 moles/ 5,00L = 0,0101M

Replacing in (1)

kc = 0,116

b. Total pressure could be obtained from the total molarity, thus:

P = MRT

Total molarity (M) is: 5,986x10⁻³M + 0,0203M + 0,0101M = 0,0364M

Thus, total pressure is:

P = 0,0364 mol/L × 0,082atmL/molK×373,15K  -100°C=373,15K- = 1,11 atm

c. Using the equilibrium it is possible to obtain initial moles of NOBr with moles of each compound in equilibrium and then, initial mass, thus:

0,101 moles NO× = 0,101 mol NOBr

0,0507 moles Br₂× = 0,101 mol NOBr

Thus, total initial moles of NOBr are:

0,101 mol + 0,101 mol + 0,0299 mol = 0,232 mol NOBr. In mass:

0,232 mol× = 25,5 g of NOBr

I hope it helps!