▾ Part A At 25°C, AH = 1.895 kJ and AS = -3.363 J/K for the transition C(graphite) C(diamond) ->> Based on these data graphite cannot be converted to diamond at 1 atm pressure. O graphite is more stable than diamond below 290°C and diamond is more stable than graphite above 290°C. diamond is more stable than graphite below 290°C and graphite is more stable than diamond above 290°C. diamond is more stable than graphite at all temperatures at 1 atm.

▾ Part A At 25°C, AH = 1.895 kJ and AS = -3.363 J/K for the transition C(graphite) C(diamond) ->> Based on these data graphite cannot be converted to diamond at 1 atm pressure. O graphite is more stable than diamond below 290°C and diamond is more stable than graphite above 290°C. diamond is more stable than graphite below 290°C and graphite is more stable than diamond above 290°C. diamond is more stable than graphite at all temperatures at 1 atm.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Similar questions

1. The reaction between hydrazine (N2H4) and dinitrogen tetroxide (N2O4) in their liquid forms is used to propel rockets into the air. 2 N2H41) + N2O40) → 3 N2(g) + 4 H2O2(g) Substance N2(a) H2O(a) S°(J/mol-K) 121.2 304.3 191.5 188.7 He (kJ/mol) 50.6 9.16 -241.8 Calculate the AH. AS and AG° for the given reactions below. Tell whether the process is spontaneous or not at standard conditions.
Calculate AS°sys for the below chemical reaction using the provided AS° values. CaCO3 (s, calcite) + 2 HCI (g) → CaCl2 (s) + CO2 (g) + H2O (1) Substance AS° (J/(mol•K)) HCI (g) 186.69 CaCl2 (s) 104.6 CaCO3 (s, calcite) 92.88 CO2 (g) 213.6 H2O (1) 69.91 H20 (g) 188.83 227.5 J/(mol•K) -78.15 J/(mol•K) -101.6 J/(mol•K) 40.77 J/(mol•K) 108.5 J/(mol•K)
Example: Calculate the AS˚ sys for 1 mol C2H5OH(I) → C2H5OH(g) • S° C2H5OH (1) = 160.70 J/mol K S° C2H5OH (g) = 282.70 J/mol K
Determine AS° for 2 03(g) → 3 O₂(g). Substance S° (J/mol.K) 03 (9) 239 O2 (g) 205 - J/mol · K
Calculate the standard free energy change (kJ/mol) for the reaction shown at a temperature of 298.15 K using the information provided. Reaction AHrxn (kJ/mol) ASrxn (J/mol-K) N2(g) + 2 02(g) → N204(g) 11.1 -297.6
Consider the following reaction: SO((g) + CaO(s) = CasO4(s). If AGº is -345 kJ/mol for this reaction at 298K, what is the Keg at this temperature? Recall that R is 8.314 J mol-1 K-1. A) 3.1 x 1040 B) 3.50 x 1050 c) 2.96 x 1060 D 2.23 x 1070
50.Calculate ΔΔS° for the following reaction, 2SO2(g) + O2(g) ⟶⟶ 2SO3(g) given S°[SO2(g)] = 248.2 J/mol×K, S°[O2(g)] = 205.1 J/mol×K, and S°[SO3(g)] = 256.8 J/mol×K. a. –196.5 J/K b. –94.0 J/K c. –187.9 J/K d. +187.9 J/K e. +196.5 J/K
Calculate the AS˚ sys for 1 mol C2H5OH(I) → C2H5OH(g) • S° C2H5OH (I) = 160.70 J/mol K S° C2H5OH (g) = 282.70 J/mol K
One gram of ice melts with absorption of 333J or energy at 0C. Is melting of ice spontaneous at -5C? at 10C?
6.) Given the information about S° for each species in the table below, calculate AS°rxn for the following reaction: Compound S°(J/mol'K) 4 NH3(g) + 5 02(g) → 4 NO(g) + 6 H2O(g) NH3(g) 192.8 205.2 02(g) NO(g) H20(g) 210.8 188.8
chapter 12: 8) Calculate the free-energy change of the following reaction at 448°C and standard pressure. Values in the table are at standard pressure and 25°C. C2H4(g)+3O2(g)--->2CO2(g)+2H2O(g) ΔHºf,(kJ/mol) Sºf, J/mol•K ΔGºf, kJ/mol C2H4(g) 52.3 219.5 68.1 O2(g) 0 205.0 0 CO2(g) –393.5 213.6 –394.4 H2O(g) –241.8