Outcomes Group A Time when reaction mixture was included to NaHCO3 (s) Volume of Na2S2O3 added (cm3) Group E Time when reaction mix was contributed to NaHCO3 (s) Volume of Na2S2O3 included (cm3) 0 Questions: 1. Compose a balanced chemical equation to represent the response in between iodine and propanone in acidic medium. 2. What is the function of the salt hydrogencarbonate? Sodium hydrogencarbonate option is utilized to quench the reaction in this experiment.
When the reaction mix is transferred into the conical flask consisting of sodium hydrogencarbonate solution, it reduces the effects of the sulphuric acid in the response mix.
2NaHCO3 + H2SO4–> > Na2SO4 + 2CO2 + 2H2O At room temperature, without the existence of hydrogen ions (driver), the rate of the response in between propanone and iodine is extremely slow and is practically stopped. 3. Discuss why the concentration of iodine in the reaction mix can be revealed in terms of the volume of sodium thiosulphate added.
In the titration, reaction between iodine and thiosulphate(VI) ion: I2(aq) + 2S2O32-(aq) i ?? S4O62-(aq) + 2I-(aq)? no. of mole of iodine in the reaction mix =(1/2)(that of sodium thiosulphate included) =(1/2) (volume of salt thiosulphate included)(molarity of salt thiosulphate added)?(no. of mole of iodine in the response mixture)/(volume of reaction mix) = [( 1/2) (volume of salt thiosulphate added)(molarity of sodium thiosulphate included)]/(volume of response mix)? concentration of iodine in the reaction mixture= [( 1/2) (volume of sodium thiosulphate included)(molarity of salt thiosulphate included)]/(volume of reaction mix)?
The concentration of iodine in the response mix can be expressed in terms of the volume of salt thiosulphate added.
4. Plot a graph of the time at which the 10cm3 samples of the reaction mix were included to the salt hydrogencarbonate option(x-axis) versus the volume of sodium thiosulphate needed to respond with the staying iodine(y-axis). Connected 5. Figure out the concentration of salt thiosulphate from the chart you outlined.
From the data of Group A ,at time=0, volume of sodium thiosulphate added should be 20. 75cm3. ?concentration of iodine in the reaction mixture = [(1/2) (volume of sodium thiosulphate added)(molarity of sodium thiosulphate added)] /(volume of reaction mixture) ?0. 0198= [(1/2) (20. 75/1000) (molarity of sodium thiosulphate added)]/(50/1000) molarity of sodium thiosulphate added =0. 095421686 ~0. 0954M 6. What is the order of reaction with respect to iodine? i. e. what is the value of n in the equation: Rate of reaction=constant[I2]n.
Throughout the experiment, the iodine concentration in the reaction fell as iodine was consumed by propanone In fact ,the slope of the straight line in the graph is the rate of equation. From the graph, the iodine concentration changed at a uniform rate throughout the experiment as the slope of the graph is constant. Thus, the rate of iodine concentration is independent on the iodine concentration and hence the reaction is zero order with respect to iodine. The order of reaction with respect to iodine is zero, i. e. n = 0 7. Does iodine take part in the rate determining step of the reaction between iodine and propanone?
Since the reaction is zero order with respect to iodine ,the iodine plays no part in the rate determining step of the reaction 8. Record the gradients of the graphs obtained by other four groups Group A B C D E Volume of propanone added (cm3) 25. 0 20. 0 15. 0 10. 0 5. 0 Gradient of graph/cm3S-1 -2. 92 x10-3 -2. 12 x10-3 -2. 17 x10-3 -2 x10-3 -1. 86 x10-3 9. Plot the gradients of graphs above against the initial volume of propanone solution added. A graph showing the gradients of graphs above against the initial volume of propanone solution added.
10. What is the order of the reaction with respect to propanone? The gradient of the graph in (9)is directly proportional to the rate of reaction and volume of propanone solution is also directly proportional to its concentration. Since a straight line is plotted in the graph above, rate of reaction is directly proportional to the concentration of propanone. Thus, the order of the reaction with respect to propanone is 1. Conclusion The order of the reaction with respect to propanone is 1 and the order of the reaction with respect to iodine is 0.
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