Enthalpy and Entropy Change in the Reaction Hi Chegg, One of the experts mention

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Enthalpy and Entropy Change in the Reaction

Hi Chegg,

One of the experts mentioned that I can post multiple set of Q&A somewhere. However, I don't quite find where i can do that. It would be nice if someone can show me how. Nevertheless, I need help on completing the following table. Please provide steps for 2 rows and I will be able to figure out the rest. Thank you.

3 M Fe(NO3)3 solution and 10 mL of 3.00x 10 solution are pipetted into a 25.00 mL volumetric flask. M SCN 1. 10.00 mL of 1.50 2. The mixture is diluted to the mark with 0.10 M HNO3 and mixed well. Sample Part B Spreadsheet for Prelab. Solution [Fe3+]e T(oc) Abs [Fe3+] init TscN linit. K 1/T(K-1) Ink [SCN leg 15.5 0.618 26.3 0.582 36.2 0.536 44.7 0.5 13 T Given that a 1.00 x 10-4 M Standard FeSCN solution has an absorbance of 0.429 at 22.5 oc Complete the rest of the table by entering correct expressions to calculate the indicated quantities.

Explanation / Answer

Wow, this is a long one (for readibility, I broke it up, and will only include the absorbance data on the one that matters -- the [FeSCN2+eq]:

T(°C) [Fe3+]init
15.5 (1.50 x 10^-3 M)*(10.0 mL)/(25.00 mL) = 9.3 x 10^-3
26.3 (1.50 x 10^-3 M)*(10.0 mL)/(25.00 mL) = 0.01578
36.2 (1.50 x 10^-3 M)*(10.0 mL)/(25.00 mL) = 0.02172
44.7 (1.50 x 10^-3 M)*(10.0 mL)/(25.00 mL) = 0.02682


T(°C) [SCN-]init
15.5 (3.00 x 10^-3 M)*(10.00 mL)/(25.00 mL) = 0.0186
26.3 (3.00 x 10^-3 M)*(10.00 mL)/(25.00 mL) = 0.03156
36.2(3.00 x 10^-3 M)*(10.00 mL)/(25.00 mL) = 0.04344
44.7 (3.00 x 10^-3 M)*(10.00 mL)/(25.00 mL) = 0.05364

T(°C) Abs [FeSCN2+]eq
15.5* 0.655 (1.00 x 10-4 M)*(0.655)/(0.434) = 1.55 x 10^-3
26.3*0.582 (1.00 x 10-4 M)*(0.582)/(0.434) = 2 x 10^-3

Similarly solve for other temperatures

T(°C) [Fe3+]eq
15.5 [(1.50 x 10-3 M)*(10.0 mL)/(25.00 mL)] – [(1.00 x 10-4 M)*(0.618)/(0.429)] = ......?
26.3 [(1.50 x 10-3 M)*(10.0 mL)/(25.00 mL)] – [(1.00 x 10-4 M)*(0.582)/(0.429)] = ........?
Similarly solve for other temperatures

T(°C) [SCN-]eq
15.5 (3.00 x 10-3 M)*(10.00 mL)/(25.00 mL) – [(1.00 x 10-4 M)*(0.618)/(0.429)] = ............?
26.3 (3.00 x 10-3 M)*(10.00 mL)/(25.00 mL) – [(1.00 x 10-4 M)*(0.582)/(0.429)] = ..............?
Similarly solve for other temperatures...

T(°C) K
15.5[Fe3+]*[SCN-]/[FeSCN2+] =...........?
26.3 [Fe3+]*[SCN-]/[FeSCN2+] = .........?
36.2 [Fe3+]*[SCN-]/[FeSCN2+] =.................?
44.7 [Fe3+]*[SCN-]/[FeSCN2+] = ...........?
substitute your values as calculated above...
T(°C) 1/T (°K^-1)
15.5* 1/(15.5 + 273.15) = 0.05369
26.3* 1/(26.3 + 273.15) = 0.0878
Similarly solve for other temperatures

T(°C) ln K
15.5 ln (0.05369) = -19.6866
Similarly solve for other temperatures

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