Physical Chemistry
2nd Edition
ISBN: 9781133958437
Author: Ball, David W. (david Warren), BAER, Tomas
Publisher: Wadsworth Cengage Learning,
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Chapter 18, Problem 18.11E
Interpretation Introduction
Interpretation:
The value of
Concept introduction:
The point at which the bond between the two atoms become nonexistent, that is, the molecule exits as two separated atoms is known as dissociation limit. The relation between
Where,
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•
•
•
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I still keep getting this question wrong. I put 4.2 E -2 as my answer, and it says I'm correct. So then, I tried 4.16 E -2, then 4.1 E-2, but it still didn't work. I am incredibly lost in this assignment. What am I doing wrong?
C+02→CO₂(g)
H₂ (g)+1/2 O2(g)→H2O(1)
C14H10(s)+33/2 O₂(g)→14CO₂(g)+5H₂O(1)
Using Hess's law combine these three reactions to obtain the net anthracene formation reaction
(14C(graphite)+5H₂(g)→C₁4H10(s)) and calculate the value of anthracene and repeat to obtain the
value of phenanthrene.
¹AHⓇ form CO₂:
2AH form H₂O:
-393.5 kJ/mol
-285.83 kJ/mol
The sun supplies energy at a rate of about 1.0 kilowatt per square meter of surface area (1 watt = 1 J/s). The plants in
an agricultural field produce the equivalent of 21 kg of sucrose (C12 H22 O11) per hour per hectare (1 ha = 10,000 m
Assuming that sucrose is produced by the reaction
m²).
12CO2 (9) + 11H2O(1)
→ C12H22O11(s) + 120O2(g) AH=5640 kJ
calculate the percentage of sunlight used to produce the sucrose – that is, determine the efficiency of photosynthesis in
this field.
Percent efficiency
Chapter 18 Solutions
Physical Chemistry
Ch. 18 - Prob. 18.1ECh. 18 - Prob. 18.2ECh. 18 - Prob. 18.3ECh. 18 - Prob. 18.4ECh. 18 - The following are the first four electronic energy...Ch. 18 - Prob. 18.6ECh. 18 - Prob. 18.7ECh. 18 - Prob. 18.8ECh. 18 - Prob. 18.9ECh. 18 - Prob. 18.10E
Ch. 18 - Prob. 18.11ECh. 18 - Prob. 18.12ECh. 18 - Prob. 18.13ECh. 18 - Prob. 18.14ECh. 18 - Prob. 18.15ECh. 18 - Prob. 18.16ECh. 18 - Prob. 18.17ECh. 18 - Prob. 18.18ECh. 18 - Prob. 18.19ECh. 18 - Prob. 18.20ECh. 18 - Prob. 18.21ECh. 18 - Prob. 18.22ECh. 18 - Prob. 18.23ECh. 18 - Prob. 18.24ECh. 18 - Prob. 18.25ECh. 18 - Prob. 18.26ECh. 18 - Prob. 18.27ECh. 18 - Prob. 18.28ECh. 18 - Prob. 18.29ECh. 18 - Prob. 18.30ECh. 18 - Prob. 18.31ECh. 18 - Prob. 18.32ECh. 18 - Prob. 18.33ECh. 18 - What are qnuc and qrot for N2(I=1)? See Table 18.3...Ch. 18 - The rovibrational spectrum of acetylene, HCCH,...Ch. 18 - Prob. 18.36ECh. 18 - Prob. 18.37ECh. 18 - Prob. 18.38ECh. 18 - Prob. 18.39ECh. 18 - Prob. 18.40ECh. 18 - Prob. 18.41ECh. 18 - Prob. 18.42ECh. 18 - Use equation 18.44 to show that pV=NkT.Ch. 18 - Prob. 18.44ECh. 18 - Determine E,H,G, and S for CH4 at standard...Ch. 18 - Prob. 18.48ECh. 18 - Prob. 18.49ECh. 18 - Calculate the heat capacity of NO2 at 298K and...Ch. 18 - Prob. 18.51ECh. 18 - In Chapters 17 and 18 we have derived expressions...Ch. 18 - Prob. 18.55ECh. 18 - Prob. 18.56ECh. 18 - Prob. 18.57ECh. 18 - Prob. 18.58ECh. 18 - Prob. 18.59ECh. 18 - Prob. 18.60E
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- Consider the following data. 2 H₂0 (1) 2 H₂(g) + O₂(g) 2 HNO3(1) = N₂05(9) + H₂O(1) 2 HNO30 (1) ⇒ N₂(g) + 3 O₂(g) + H₂(g) Use Hess's law to calculate AH for the reaction below. ΔΗ= |1083.8 AH = +571.7 kJ AH = +92.0 kJ AH = +348.2 kJ x kJ 2 N₂05(9) 2 N₂(g) + 5 O₂(g)arrow_forwardThe sun supplies energy at a rate of about 1.0 kilowatt per square meter of surface area (1 watt = 1 J/s). The plants in an agricultural field produce the equivalent of 23 kg of sucrose (C12H22 O11) per hour per hectare (1 ha = 10,000 m²). Assuming that sucrose is produced by the reaction 12CO₂(g) + 11H₂O(l) → C12H22 O11 (8) + 1202 (9) AH = 5640 kJ calculate the percentage of sunlight used to produce the sucrose - that is, determine the efficiency of photosynthesis in this field. Percent efficiency = %arrow_forwardThe following information is given for antimony at 1atm: Th = 1440.00°C ΔΗ, vap (1440.00°C) = 1.605 x 10³ J/g Tm = 631.00°C Specific heat solid = 0.2090 J/g °C Specific heat liquid = 0.2590 J/g °C AHfus (631.00°C) = 161.1 J/g A 40.40 g sample of liquid antimony at 721.00°C is poured into a mold and allowed to cool to 25.00°C. How many kJ of energy are released in this process? (Report the answer as a positive number.) Energy = kJarrow_forward
- The following information is given for antimony at latm: AHvap (1440.00°C) = 1.605 x 10° J/g AĦfus (631.00°C) = 161.1 J/g Th = 1440.00°C %3D %3D Tm = 631.00°C Specific heat solid = 0.2090 J/g °C Specific heat liquid = 0.2590 J/g °C A 31.40 g sample of liquid antimony at 738.00°C is poured into a mold and allowed to cool to 22.00°C. How many kJ of energy are released in this process? (Report the answer as a positive number.) Energy = kJarrow_forward6. The plot given below was constructed from data that was collected for the following reaction. N2(0) + 3 H2(g) = 2 NH3(9) = 2 NH3(g) Using the information provided in the plot, find AH°, AS°, AG°, and Keq for the given reaction at 25 °C. Van't Hoff Plot y = 5556.9x - 11.92 8. 4 2 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035 0.004 0.0045 1/T (1/K) 12 10arrow_forwardThank you!arrow_forward
- The following information is given for antimony at 1 atm: Tb= 1440.00°C Hvsp=(1440.00°C) = 1.605*10^3 J/g Tm= 631.00°C H fus=(631.00°C) =161.1 J/g Specific heat solid = 0.2090 J/g °C Specific heat liquid = 0.2590 J/g °C A 21.10 g sample of solid antimony is initially at 609.00°C. If the sample is heated at constant pressure ( P= 1 atm), what kJ of heat is needed to raise the temperature of the sample to 729.00°C.arrow_forwardThe sun supplies energy at a rate of about 1.0 kilowatt per square meter of surface area (1 watt =1 J/s). The plants in an agricultural field produce the equivalent of 13 kg of sucrose (C12 H22 O11) per hour per hectare (1 ha : 10,000 m²). Assuming that sucrose is produced by the reaction 12CO2 (g) + 11H2O(1) → C12 H22 O11 (s) + 1202 (g) AH=5640 kJ calculate the percentage of sunlight used to produce the sucrose that is, determine the efficiency of photosynthesis in this field. Percent efficiency %arrow_forwardThe potential energy associated with two charged particles is 34 mJ when they are separated by a distance d. What is the potential energy associated with them (in mJ) when they are separated by 3d?arrow_forward
- given with the correct answer by my professor, but there was no explanation on how this correct or show the steps to getting the answer. Could you please help me answer this with explanation of this problem? please and thank you.arrow_forwardYou are given the following data: 2H(g) > H₂(g) - 2Br(g) - Br₂ (g) 2HBr(g) → H₂(g) + Br₂ (g) kJ mol → x10 ΔΗ° = -436.4 Calculate AH° for the reaction. H(g) + Br(g) HBr (g) Be sure your answer has the correct number of significant digits. X ΔΗ° = -192.5 ΔΗ° = 72.4 kJ mol kJ mol kJ molarrow_forwardConstants: • c = 2.9979 x 108 m/s h = 6.626 x 10-34 J s per one photon • R = 8.314 J/(K mol) = 0.08206 L atm/(K mol) NA = 6.022 x 1023 particles/mol RH = 1.097 x 107 m1 = 2.178 x 10-18 J . . You have an aqueous Glucose solution that is 12.0% Glucose by mass. What is the molality of Glucose in the solution? Molar Mass of Water = 18.015 g/mol Molar Mass of Glucose 180.16 g/mol =arrow_forward
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