Answer to Question #305190 in Organic Chemistry for claire yeboah

Question

1-Bromobutane was hydrolysed by an aqueous sodium hydroxide (NaOH) solution. 

 

     CH₃CH₂CH₂CH₂Br    +   NaOH     →      CH₃CH₂CH₂CH₂OH      +     NaCl 

        1-Bromobutane                                             Butan-1-ol

 

                       During the reaction, the reactant sodium hydroxide is used up. Samples of the reaction mixture were drawn off at regular time intervals and analysed by titration with standard sulfuric acid to find out the concentration of sodium hydroxide. 

      The results are given in the table below:

 

 

 

 Time / s Concentration of OH-  / mol dm-3

0 0.500

100 0.350

200 0.250

300 0.180

400 0.125

500 0.090

600 0.063

700 0.040

800 0.030

 Select 4 suitable points along your graph [see part (a)] and draw tangents to find the rates. Construct a table of “Rate” against “Concentration” (2 marks). The rates need to be calculated and the calculations shown 

Answer 1

Calculating the rate of reaction from the gradient of a tangent

To calculate the rate at the start of a reaction, you need to find the mathematical gradient (steepness) of the gradient. Use this equation:

rate of reaction=change in mass or volume of productchange in time

Volumes of solutions in reactionsCalculating amounts from concentration and volume

The amount in be calculated if data in real the solution is known. The mass of solute can be calculated

Concentration in mol/dm3=amount of solute in molvolume in dm3

Gives:

Amount of solute in mol = concentration in mol/dm³ × volume in dm³

Calculate the amount of sodium hydroxide, NaOH, in 25.0 cm³ of solution of concentration 0.1 mol/dm³.

Converting the volume from cm³ to dm³, 25.0 cm³ = 25.0 ÷ 1000 = 0.025 dm³.

Amount of NaOH in mol = concentration in mol/dm³ × volume in dm³

= 0.1 mol/dm³ × 0.025 dm³

= 0.0025 mol

Calculating masses from concentration and volume

If the amount in mol of a solute in a given volume of solution is known, its mass can also be calculated.

Question

For the example above, calculate the mass of NaOH in 25.0 cm³ of solution. (M_r = 40 g)

Reveal answer

More concentration calculations

If the volumes of two solutions that react completely are known, and the concentration of one of the solutions is known, then the concentration of the other solution can be calculated.

Example

25.0 cm³ of hydrochloric acid, HCl, of concentration 1.0 mol/dm³ reacts with 20.0 cm³ of sodium hydroxide, NaOH, solution. Calculate the concentration of the NaOH solution. The equation for the reaction is HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

Converting volumes from cm³ to dm³:

• volume of HCl = 25.0 cm³ = 25.0 ÷ 1000 = 0.025 dm³
• volume of NaOH = 20.0 cm³ = 20.0 ÷ 1000 = 0.020 dm³

Substituting into the equation to find the amount of HCl in the given volume:

Amount of HCl in mol = concentration in mol/dm³ × volume in dm³

= 1.0 mol/dm³ × 0.025 dm³

= 0.025 mol

The equation shows that 1 mol of HCl reacts with 1 mol of NaOH. So the amount of NaOH in 20.0 cm³ is 0.025 mol.

Substituting into the equation to find the concentration of NaOH:

Concentration of NaOH/dm3=amount of solute in molvolume in dm3

Concentration of NaOH/dm3=0.025 mol0.020 dm3

= 1.25 mol/dm³

The total given concentration of hydroxyl ion will be calculated by following this procedure( 0.50+1000.350+2000.250+3000+180+4000.125+5000.090+6000+063+7000.040+8000.030) Mol/dm³

= 36​001.628Mol/dm³