C6H14(l) + Br2(l) + light → C6H13Br(l) + HBr(g) How do I explain this reaction?

Expert Solution

the reaction is a radical substitution reaction. it is also a chain reaction(self propagating reaction)

Brown bromine disappears, while hydrogen bromide and bromohexane are formed:

C6H14(l) + Br2(l) -----light----> C6H13Br(l) + HBr(g)

The only observable change you will be a change in colour. Bromohexane which is the final product of the reaction is colourless. so as the reaction proceeds there will be gradual loss of the reddish brown colour until the solution turns colourless.

a completely colourless solution will result if the the Hexane does not run out before the completion of the reaction. If the amount of hexane is limited the end product will not be completely colourless as it will contain excess bromine*( which is responsible for the reddish brown colouration).

in other words a totally colourless solution will result if all the bromine is consumed in the course of the reaction.

queen_honey_blossom answered 4 months ago

8. The equilibrium constant Kc for the reaction H2(g) + Br2(g) ⇌ 2 HBr(g) is 2.18×106...

8. The equilibrium constant Kc for the reaction H2(g) + Br2(g) ⇌ 2 HBr(g) is 2.18×106 at 730°C. Starting 3.20 moles of HBr in a 12.0-L reaction vessel, calculate the concentrations of H2, Br2, and HBr at equilibrium. the answer is [H2] = [Br2] = 1.81×10-4 M [HBr] = 0.267 M but how and why?

2Na(s) + Br2= 2NaBr A reaction of 46.1 g of Na and 53.1 g of Br2...

2Na(s) + Br2= 2NaBr A reaction of 46.1 g of Na and 53.1 g of Br2 yields 56.7 g of NaBr. What is the percent yield?

For the reaction show below determine the rate law. CH3COCH3 + Br2 → CH3COCH2Br + HBr(aq)...

For the reaction show below determine the rate law. CH3COCH3 + Br2 → CH3COCH2Br + HBr(aq) The bromination of acetone, an organic solvent, is catalyzed by an acid. The rate of disappearance of bromine was measured for several different concentrations of the reactants including the catalyst, H+. Therefore, the catalyst will appear in the rate law, but not in the overall reaction. The following data were obtained: [CH3COCH3] [Br2] [H+] Initial Rate (mol/L s) 0.30 0.05 0.05 5.7 x 10-5...

A reaction of 59.5 g of Na and 48.7 g of Br2 yields 57.1 g of...

A reaction of 59.5 g of Na and 48.7 g of Br2 yields 57.1 g of NaBr. What is the percent yield?

1. The decomposition reaction of NOBr(g) to 2 NO(g) and Br2(g) is second order with k...

1. The decomposition reaction of NOBr(g) to 2 NO(g) and Br2(g) is second order with k = 0.90 M-1∙s-1. 2 NOBr(g) …….> 2 NO(g) + Br2(g) What is the half-life of this reaction if the initial concentration of NOBr is 0.78 M? 2. The decomposition reaction of NOBr(g) to 2 NO(g) and Br2(g) is second order with k = 0.90 M-1∙s-1. 2 NOBr(g) …….> 2 NO(g) + Br2(g) How much time would pass before a 4.00 g sample of NOBr...

Problem 1: For the reaction 2CH3OH(l) + 3O2(g) = 2CO2 + 4H2O(l) I calculated that the...

Problem 1: For the reaction 2CH3OH(l) + 3O2(g) = 2CO2 + 4H2O(l) I calculated that the change in molar entropy of the system is 313.7 Joules/(moles*kelvin). The question then asked what is the entropy change of the system when 2.35 moles of O2 react. I multiplied 313.7 Joules/(moles*kelvin) by 2.35 moles of O2 and also divided by 3 moles of O2 because of the "3" coefficient, and got 246 Joules/kelvin and I just want to make sure that last part...

The equilibrium constant for the following reaction: H2(g) + Br2(g) ↔ 2HBr (g) is 1.95 x...

The equilibrium constant for the following reaction: H2(g) + Br2(g) ↔ 2HBr (g) is 1.95 x 103 at a certain temperature. Find the equilibrium pressure of HBr if 10.70 atm of HBr is introduced into a sealed container at this temperature.

For the reaction Br2 (g) + Cl2 (g) → 2 BrCl Kp = 1.11 x 10-4...

For the reaction Br2 (g) + Cl2 (g) → 2 BrCl Kp = 1.11 x 10-4 at 150K. a) If 1.0 atm each of bromine and chlorine gas are placed in an otherwise empty 1 L container and allowed to equilibrate at a temperature of 150K, what will be the partial pressures at equilibrium of all three gases? Br2 atm Cl2 atm BrCl atm b) suppose 0.2 atm each of all three gases are placed in a selaed container and allowed to equilibrate...

The reaction H2(g) + Br2(g) 2HBr(g) has Kc = 2.0 x 109 at 25 °C. If...

The reaction H2(g) + Br2(g) 2HBr(g) has Kc = 2.0 x 109 at 25 °C. If 0.100 mol of H2 and 0.100 mol of Br2 were placed in a 10.0 L container and allowed to react, what would all the equilibrium concentrations be at 25 °C?

The equilibrium constant Kc for the reaction below is 0.00427 at a certain temperature. Br2(g) ⇌...

The equilibrium constant Kc for the reaction below is 0.00427 at a certain temperature. Br2(g) ⇌ 2Br(g) If the initial concentrations are [Br2] = 0.0844 M and [Br] = 0.0763 M, calculate the concentrations of these species at equilibrium.

Question