Need help with question 2 please? ratory Exercise: Rates of Chemical Reactions f

Question

Need help with question 2 please? ratory Exercise: Rates of Chemical Reactions four factors that affect reaction rate and explain how they affect the rate Co) lis, o.is.onen ame four +ne 0 em netur factors that affect reaction rate kuee in owtheymdthdae mon Caaster and moxe colisions occur ) 2. A student studied the clock reaction described below and observed the following data: Initial Concentrations (M) Reaction Time Relative Rate S.O seconds 150 Rxn #1 Rxn #2 0.060 0.060 0.120 39 After mixing the following reagents for reaction #2, the reaction time was recorded to be 78 seconds. Reaction Mixture #2:15 mL 0.22 M KI 25 mL 0.13 M Na,S,O, 15 mL 0.0010 M Na,S,O, a. Find the concentration of the following reagents in the mixture. (M,V, M,V, 11 00865 .8 se What is the relative rate of reaction (1000/t)? b. b. What is the relative ruts of reaction (10000? c. Fill in the table with the missing data. continued on the back Appendix 33

Explanation / Answer

Solution :-

The concentrations for the species in the trial 2 are calculated as follows

After mixing the solutions total volume is = 15.0 ml + 25.0 ml + 15.0 ml = 55.0 ml

There fore the concentration of the each species is as follows

[ I^- ] = 0.22 M * 15 ml / 55 ml = 0.06

[S2O8^2-] = 0.13 M * 25 ml / 55 ml = 0.0591 M

[S2O3^2-] = 0.0010 M * 15.0 ml / 55 ml = 0.000273 M

These concentration values goes into the table for the trial 2

using the data from the table need to solve the Q2

Now Q2

a) Calculating the exponent of the S2O8^2-

To calculate the exponent we need to use the concentrations and rate

Rate 2 / rate 1 = ([S2O8^2-]2/[S2O8^2-]1)^x

12.82/6.666 = ([0.06]/[0.027])^x

1.92 = 2.22^x

Log 1.92 = x * log 2.22

X= log 1.92 / log 2.22

X = 0.82 rounded to 1

Therefore the exponent of the S2O8^2- is 1

That is reaction is first order with the S2O8^2-

b) Now lets find the exponent for the I^-

Rate 3 /rate 2 = ([I^-]3/[I^-]2]^y

25.64/12.82 = (0.120 / 0.060)^y

2 = 2^y

Log 2 = y* log 2

Y = log 2 / log 2

Y= 1

Therefore the exponent for the I^- is also 1

Therefore reaction is forst order with the I^-

Therefore the rate equation for the reaction is as follows

Rate = K [S2O8^2-][I^-]

c) Now lets find the overall order

Overall order of reaction is the sum of the orders of the individual reactants

Therefore

Overall order = 1 +1 = 2

So the overall order of the reaction is 2

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