Question

nonequivalent hydrogens on adjacent carbons will show multiple peaks on the NMR, a phenomena known as

Answer 1

Answer on Question #63427, Chemistry / General Chemistry

Chapter 15 (15.74)

A sample of nitrosyl bromide (NOBr) decomposes according to the equation $2\mathrm{NOBr}(\mathrm{g}) \rightleftharpoons 2\mathrm{NO}(\mathrm{g}) + \mathrm{Br}2(\mathrm{g})$

An equilibrium mixture in a 5.00-L vessel at $100^{\circ}\mathrm{C}$ contains $3.27\mathrm{g}$ of NOBr, $3.09\mathrm{g}$ of NO, and $8.23\mathrm{g}$ 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?

Solution:

1) Moles = Mass/Mol.mass

Moles NOBr = 3.27 g / 109.8 g/mol = 0.0298 mol NOBr

Moles NO = 3.09 g / 29.9 g/mol = 0.1033 mol NO

Moles Br2 = 8.23 g / 159.9 g/mol = 0.0515 mol Br2

Concentration of NOBr = 0.0298 mol / 5.00 L = 0.00596 mol / L

Concentration of NO = 0.1033 mol / 5.00 L = 0.02066 mol / L

Concentration of Br2 = 0.0515 mol / 5.00 L = 0.0103 mol / L

Kc for the equation $= [\mathrm{NO}]^{\wedge}2^{*}[\mathrm{Br}2] / [\mathrm{NOBr}]^{\wedge}2$

Kc = [0.02066 mol / L]^2 * [0.0103 mol / L] / [0.00596 mol / L]^2 = 17.6 x 10^-3 mol / L

2) Use the ideal gas equation $\mathrm{PV} = \mathrm{nRT}$

P=nRT/V

n= 0.0298 mol + 0.1033 mol + 0.0515 mol = 0.1846 mol

T = 100 + 273 = 373 K

P = 0.1846 mol x 373 K x 8.3 J/molK / 0.005 m³ = 114300.6 Pa = 857.3 Tor

3) We use conservation of mass

m (NOBr) = 3.27 g + 3.09 g + 8.23 g = 14.59 g

Answer: 1) $17.6 \times 10^{-3} \, \text{mol/L}$ ; 2) 857.3 Tor; 3) $14.59 \, \text{g}$