wet lab 4.pdf

# wet lab 4.pdf - Show a complete sample calculation of how...

• Lab Report
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Show a complete sample calculation of how you calculated the initial I - concentration [I - ] = 0.200 M V(I - ) = 10.0 mL = 0.0100 L V total = 50.0mL = 0.0500 L [I - ] initial = ? Use dilution equation: C 1 V 1 = C 2 V 2 In this case: [I - ]xV(I - ) = [I - ] initial x V total [I - ] initial = ([I - ]xV(I - ))÷(V total ) = (0.200 M x 0.0100 L) ÷ (0.0500 L) = 0.0400 M

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Show a complete sample calculation of how you calculated the rate of reaction Firstly, the n(S 2 O 8 2- ) needs to be determined by using mole ratio. We know that, n(S 2 O 3 2- ) = [S 2 O 3 2- ]xV(S 2 O 3 2- ) = (5.00x10 -3 mol/L) x (0.01000 L) = 5.00 x 10 -5 mol Therefore, n(S 2 O 8 2- ) equals: 5.00x10 -5 mol S 2 O 3 2- x (1 mol I 3 - )/(2 mol S 2 O 3 2- ) x (1 mol S 2 O 8 2- )/(1 mol I 3 - ) n(S 2 O 8 2- ) consumed = 2.50 x 10 -5 mol S 2 O 8 2- V total of reaction = 50.0 mL = 0.0500 L Hence, decrease in [S 2 O 8 2- ] equals: [S 2 O 8 2- ] = n(S 2 O 8 2- ) consumed ÷ V total of reaction = 2.50 x 10 -5 mol ÷ 0.0500 L = 5.00x10 -4 M Rate = [S 2 O 8 2- ] ÷ time(s) = 5.00x10 -4 M ÷ 97 s = 5.15x10 -6 M/s

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1. Using your data from Part A, what can you conclude about the dependence of the rate on [I - ] The rate law for the reaction is: Rate = k[S 2 O 8 2- ] m [I - ] n By comparing rate laws from separate experiments with consistent [S 2 O 8 2- ] but varying [I - ], we can divide out the k constant and the [S 2 O 8 2- ]. This will allow us to compare the rate ratio to the concentration ratio [I - ] and determine the relationship between them.

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• Spring '12
• deLaat
• Chemistry, Reaction, Chemical reaction, Rate equation, Reaction rate constant

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