Experiment 2: Rate Determination and Activation Energy Introduction An important

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Experiment 2: Rate Determination and Activation Energy Introduction An important part of the kinetic analysis of energy, Ea. Activation spontaneous chemical reaction so that it will energy. An example and oxysen is spontaneous, but you need to initlate the combustion by adding activation energy from a lit match. a chemical reaction is to determine the activation energy can be defined as the energy necessary to initiate an otherwise continue to react without the need for additional of activation energy is the combustion of paper. The reaction of cellulose In this experiment you will investigate the reaction of crystal violet with sodium hydroxide. Crystal violet, in aqueous solution, is often used as an indicator in reaction of this organic molecule with sodium hydroxide can be simplified by abbreviating the chemical formula for crystal violet as CV. biochemical testing. The CV (aq) + OH' (aq) COH (aq) As the reaction proceeds, the violet-colored CV reactant will slowly change to a colorless product, following the typical behavior of an indicator. You will measure the color change with a Vernier Colorimeter or a Vernier Spectrometer. You can assume that absorbance is proportional to the concentration of crystal violet according to Beer's law. directly The molar concentration of the sodium hydroxide, NaOH, solution will be much greater than the concentration of crystal violet. This ensures that the reaction, which is first order with respect to crystal violet, will be first order overall (with respect to all reactants) throughout the experiment. You will monitor the reaction at different temperatures, while keeping the initial concentrations of the reactants the same for each trial. In this way, you will observe and measure the effect of temperature change on the rate of the reaction. From this information you will be able to calculate the activation energy,Ea, or the reaction.

Explanation / Answer

Arhenius equation is lnK= lnA- Ea/RT

with the given data points, the best fit is ( rate constant for 19 deg.c is increasing compared to 24 deg.c)

the equation is lnK= -1458/T +0.334

Ea/R= 1458 and lnA=0.334

Ea =activation energy =1458*8.314=12121 joules/mole and A= 1.39

at 14 deg.c the rate =0.004029 and at 24 deg.c the rate = 0.004935

ratto of rate constants = 0.004935/0.004020= 1.22. Hence the rule that for every 10 deg.c rise in temperature the rate is not doubled in this case.

at T1= 19 deg.c= 10+273.15= 283.15, the rate cosntant K1= 0.005731 and at T2= 40 deg.c =313.15K , let K2= rate constant ln (K2/K1)= (1458)*1/283.15-1/313.15)

K2 = 1.64*0.005731=0.0093/sec

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