# Answers For Homework 2-2 In Chemistry

(a) ΔH = (1 mole)(-487 kJ/mole) - (1 mole)(-238.6 kJ/mole) - (1 mole)(-110.5 kJ/mole) = **-138 kJ**

ΔH is negative which suggests that the reaction goes from high energy to low energy and that according to energy the reaction should be spontaneous.

ΔS = (1)(159.8 J/K) - (1)(126.8 J/K) - (1)(197.9 J/K) = **-165 J/K**

ΔS is negative which suggests that it is becoming more ordered and that according to entropy the reaction should not be spontaneous.

ΔG = (1 mole)(392.4 kJ/mole) - (1 mole)(-166.23 kJ/mole) - (1 mole)(-137.2 kJ/mole) = **-89 kJ**

ΔG is negative which suggests that the reaction should be spontaneous. In this case the energy has more influence than the entropy.

You could also calculate ΔG as follows:

ΔG = ΔH - TΔS = (-138 kJ) - (298 K)(-0.165 kJ/K) = **-89 kJ**

(b) At equilibrium ΔG = 0 = ΔH - TΔS

⇒ T = ΔH/ΔS = (-138 kJ)/(-0.165 kJ/K) = **836 K**

Below 836 K the reaction is spontaneous. Above 836 K the reaction is not spontaneous.

**Unformatted text preview: **Chemistry 329 Page 1/5 HOMEWORK #2 Answer Keys 1) You used acid‐base titrations to determine the mass percentage of KHP in an unknown sample, four analyses yielded the following results (%): 16.35, 16.25, 16.06, and 16.21 a) How should you represent this KHP mass percentage using statistical analysis? X = 16.218 S = 0.12 (or 0.12) % KHP = (16.22 ± 0.12)% or (16.22 ± 0.12)% b) Please represent this result again using confidence interval at 90% confidence level t = 2.353 μ = X ± μ = (16.22 ± 0.14)% or (16.22 ± 0.14)% 2.353(0.12 ) ts = 16.22 ± N 4 c) Using simple words, briefly explain the meaning of the representation of result in (b); There is 90% probability that the true value of the KHP mass percentage will fall into the range of (16.22 ± 0.14)% (or 16.08% ‐ 16.36%). d) You carried out another analysis, the result, 16.93%, appears to be quite different from the others. Please use statistical analysis to decide if you should keep or discard this result. Use the Grubbs test (G test) N = 5 (including the new datum!), x = 16.36, S = 0.335 G = (16.93 – 16.36)/ 0.335 = 1.70 Gtable = 1.672 for 5 observations and 95% confidence level G > Gtable therefore should discard this result (95% confidence ‐‐ there is more than a 5% chance that this new result is not a member of the same population as the other results) 2) The composition of a flake of paint found on the clothes of the victim of a hit‐and‐run accident was compared with that of paint from the car suspected of causing the accident. The following data from the spectrophotometric analysis of iron in the paint were obtained: Paint from clothes % Fe: 4.0, 4.6, 4.8, 4.5 Paint from car % Fe: 4.5, 5.3, 5.5, 5.5, 4.9 Please use statistical analyses to decide if iron compositions in the paint samples taken from the car and the victim’s clothes have a significant difference and at which confidence level. Chemistry 329 Page 2/5 X 1 = 4.475 S1 = 0.34 N1 = 4 X 2 = 5.140 S2 = 0.43 N2 = 5 Sspooled = tcalculated = (0.340 ) 2 × 3 + (0.434 ) 2 × 4 = 0.396 4+5−2 14.475 − 5.1401 4 × 5 = 2.50 0.396 4+5 For degree of freedom 4 + 5 – 2 = 7, t = 2.365 for 95% confidence level but t = 2.998 for 98% confidence level, Therefore the two results are significantly different for up to 99.5% confidence level. 3) At the wavelength of 520 nm, a solution containing 4.48 ppm KMnO4 has a transmittance of 0.309 in a 1.00 cm cell. Calculate the molar aborptivity of KMnO4. Need to first convert ppm to molarity (density of 1.00 g/mL assumed since it was not stated otherwise) 4.48mg 1mg 158.0g mol x 1000mg 1000g 1L 1.00g mL x 1000mL =2.835 x 10-5 M T = 0.309 ⇒ A = ‐logT = 0.510 Using Beer’s Law: A = εbc ⇒ ε A 0.510 = =1.799 x 104 (M -1cm-1 ) or 1.80 x 104 (M‐1cm‐1) -5 bc 1.00 x 2.833 x 10 b) The absorbance of species X at 0.0100 M concentration in a 1.00 cm cell is 0.23 at 520 nm. What is the transmittance when this solution is mixed with the KMnO4 solution in part (a) at equal volume and measured in the 1.00 cm cell? (You can neglect the error in volumes.) When mixed at equal volumes, the concentrations of both KMnO4 and species X solution are reduced to half. Since absorbance is additive, while ε and b stay the same (1.00 cm for all): A total = A KMn O4 + A x = T = 10 − A total 0.510 0.23 + = 0.255 + 0.115 = 0.37 2 2 = 10‐0.37 = 0.427 or 0.43 4) You need to determine the concentration of compounds A and B in a mixture. You have used a calibration curve to determine the absorptivities of the individual components at two different wavelengths as the following: Absorptivity (ppm‐1 cm‐1) λ (nm) A B Chemistry 329 Page 3/5 460 480 0.051 0.023 0.024 0.061 The absorbance of the mixture at 460 and 480 nm is 1.245 and 1.565, respectively. The path length is 2.00 cm. What are the concentrations of compound A and compound B in the mixture? 460 A460 = ε 460bc A + ε b bcb A 480 A480 = ε 480bc A + ε b bcb A 1.245 = 0.051x 2 xc A + 2.024 x 2 xcB 1.565 = 0.023 x 2 xc A + 0.061x 2 xcB Solve for the concentrations cB = 10 ppm c A = 7.5 ppm 5) The first few parts of this problem will help you fill in the table given below. The last few parts of this problem will have you interpret the table. Throughout this problem, be careful about using transmittance in the fractional form vs. the % form. Absorbance % Transmittance %eA(I) %eA(II) %eA(tot.) 0.001 99.77 4.4×101 4.3×102 4.4×102 1 0.010 97.72 4.4 4.3×10 4.4×101 0.100 79.43 0.55 4.3 4.4 0.300 50.12 0.29 1.5 1.5 0.600 25.12 0.29 0.72 0.78 1.000 10.0 0.43 0.43 0.61 1.500 3.162 0.92 0.29 0.96 2.500 0.3162 5.5 0.17 5.5 a) Calculate the % Transmittance for each absorbance value. A = ‐log(T). See table above. b) One of the three major sources of random error in absorbance is from the limited precision in the % transmittance, which is a constant. For most spectrophotometers, this corresponds to eT = 0.1% (absolute, not relative). Calculate the relative uncertainty for each absorbance value in the table and place these in the %eA(I) column. Note: eA = 0.43429*(eT/T). Calculate the eA using the given equation, then %eA(I)= eA/A ×100%. See table above for values. c) Another source of random error in absorbance is from the repositioning (insertion/removal) of cuvets during a set of measurements. For most spectrophotometers, this corresponds to %eT = 1% (relative, not absolute). Calculate the relative uncertainty for each absorbance value in the table and place these in the %eA(II) column. %eA(II)= eA/A ×100% = 0.43429 *(% eT) /A. See table above for values. Chemistry 329 Page 4/5 d) Since these two sources of random error are independent from each other, we can calculate the total relative uncertainty in the absorbance by the usual formula: %eA(tot.)2 = %eA(I)2 + table. %eA(II)2. Calculate these values and place them in the See table above for values. e) %eA(tot.) should be a minimum value in the approximate range of 0.400‐1.000. What is the limiting source of uncertainty for measurements made below this range or above this range? For low absorbance values, the limiting factor is the ability to reposition the cuvet on repetitive measurements. For high absorbance values, the limiting factor is the limited precision of the value of the transmittance. f) You are performing a spectrophotometric analysis and the absorbance of the unknown is ~2.300. For high precision work, a friend suggests that you place a cuvet in the spectrophotometer once and leave it in place for all measurements. This means you must use a syringe, or some other means to empty/fill the cuvet. What effect (if any) will the friend’s suggestion have on the precision of the measurements? Explain. The friend’s suggestion will not affect the precision of the final result. The suggestion will reduce the cell repositioning effect. However, at high absorbance values, the limiting factor on the precision of the measurements is in the limited precision on the transmittance measurement, not the repositioning of the cell. 7) You prepare several standards for a calibration curve and acquire the following data: Sample Absorbance Blank 0.000 2.00 ppm 0.090 4.00 ppm 0.178 6.00 ppm 0.262 8.00 ppm 0.344 10.0 ppm 0.422 12.0 ppm 0.496 14.0 ppm 0.565 16.0 ppm 0.629 a) Plot the data (absorbance vs. concentration) using a spreadsheet, and perform three different fits to the data (given below) and obtain both the equation for the line as well as the R2 coefficient. Make sure the coefficients in the equations have at least 4 sig figs. Below are the fits you should perform: 1) A straight line 2) A straight line with the intercept fixed at 0 3) A quadratic Chemistry 329 Page 5/5 0.700 y = 3.948E-02x + 1.598E-02 R2 = 9.972E-01 Absorbance 0.600 y = 4.088E-02x R2 = 9.954E-01 0.500 0.400 0.300 y = -4.608E-04x2 + 4.685E-02x - 1.224E-03 R2 = 1.000E+00 Straight line 0.200 Straight line (intercept=0) Quadratic fit 0.100 0.000 0.00E+00 4.00E+00 8.00E+00 1.20E+01 1.60E+01 Concentration (ppm) c) Three unknown samples have absorbances of 0.125, 0.510, and 0.982. What are the concentrations of these samples? Provide a brief explanation of your analysis. The quadratic is the best fit to the data set as a whole due to the way the data curve away from linearity, showing a characteristic deviation from Beer’s Law. Using this fit the concentrations of the unknowns can be calculated as 2.77 ppm and 12.4 ppm for the first two unknowns, respectively. The concentration of the last unknown cannot be determined because it extends beyond the calibration. ...

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