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# Answer to Question #163768 in Inorganic Chemistry for Nawid Sayed

Question #163768

1-Chloropropane was hydrolysed by an aqueous potassium hydroxide solution.

C3H7Cl    +   KOH     →      C3H7OH      +     KCl

1-Chloropropane                        Propan-1-ol

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

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

From the above information, Plot a Concentration of OH- (mol dm-3) against Time (s) graph. The graph MUST be a hand drawn (plotted) graph.

(b)

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

(c)

Plot a graph of Rate against Concentration of OH-. The graph MUST be a hand drawn (plotted) graph.

(d)

From your graph in part (c) deduce the Order of Reaction with respect to OH- (hydroxide ions).

(e)

Write a rate equation with respect to concentration of hydroxide ions.

(f)

Calculate the rate constant k from the gradient of the graph, drawn in part (c). Show and explain your working.

1
2021-02-15T05:56:26-0500

Rate = slope =y2 -y1/x2 -x1

= 0.175 -0 /830-200

= 0.00028moldm-3/s

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