Enter your answer in the provided box. Diamond and graphite are two crystalline forms of carbon. At 1 atm and 25°C, dia...
Diamon and graphite are two crystalline forms of carbon. At 1 atm and 25°C, diamond changes to graphite so slowly that the enthalpy change of the process must be obtained indirectly. Determine delta H rxn for C(diamond) > C(graphite) with equations from the following list: Diamond and graphite are two crystalline forms of carbon. At 1 atm and 25°C, diamond changes to graphite so slowly that the enthalpy change of the process must be obtained indirectly. Determine AHxn for C(diamond)...
1 out of 10 attemp Enter your ans wer in the provided box Diamond and graphite are two crystalline forms of carbon. At 1 atm and 25°C, diamond changes to graphite so slowly that the enthalpy change of the process must be obtained indirectly. DetermineA for rxn (diamond)- → C (graphite) C with equations from the following list: (1) Cdia (2) 2CO2(g)→2CO(g ) + O2(g) (3) C(graphite) +02(g)→CO2(g) (4) 2C0(g)→C(graphite) + CO2(g) AH AH-566.0 kJ AH--393.5 kJ AH--172.5kJ =-395.4 kJ...
carbon. Determine the AHrxn for C (diamond) → C (graphite), using the following thermochemical equations: a) C (diamond) + 02 (9) → CO2 (9) AH = -395.4 kJ b) CO (9) + 12 02 (9) → CO2 (9) AH = -283.0 kJ c) C (graphite) + CO2 (9) ► 2 CO (9) AH = +172.5 kJ
5. Use the following data to determine the delta H for the conversion of diamond into graphite: C(diamind) (s) + O2(g) = CO2(g) delta H degrees = -395.4 kJ 2CO2(g) = 2 CO(g) + O2(g) delta H degrees = 566.0 kJ 2CO(g) = C(graphite) (s) + CO2 (g) delta H degrees = -172.5 kJ C(diamond) (s) = C(graphite) delta H degrees = ?
19. Calculate AH° for the conversion of graphite into diamond from the following thermochemical equations. C(s, graphite) + O2(g) — C(s, diamond) + O2(g) CO2(g) AH° = -393.5 kJ + CO2(g) AH' = -395.9 kJ e uso -395.9 kJ +393.5 kJ -789.4 kJ +2.4 kJ -2.4 kJ
Enter your answer in the provided box From the following data, C(graphite) + O2(0)+ CO2(g) An° . =-393.5 kJ/mol rxn Hy(@) +0,6) H200 AH =-285.8 kJ/mol rxn 2C2H6(8) + 1026) →40026) + 6H20(1) AH =-3119.6 kJ/mol rxn calculate the enthalpy change for the reaction below: 2 C(graphite) + 3H2(g) → CH()
4. You are given the following information. C (graphite) + O2(g) — CO2 (g) CO(g) + 02 (g) - CO2 (g) Determine the standard enthalpy change for the reaction: C (graphite) + / O2 (g) - CO (g) AH° = - 393.5 kJ AH° = - 283.0 kJ AHrxn = ?
please help 4. You are given the following information. C (graphite) + O2 (g) - CO2 (g) CO(g) + 02 (9) - CO2 (g) Determine the standard enthalpy change for the reaction: C (graphite) + 02 (g) - CO(g) AH = - 393.5 kJ AH° = -283.0 kJ AHrxn = ?
3. (25 pts) Buckminsterfullerene is a molecule with the formula C. (60 atoms of carbon held together by covalent bonds). It was manufactured in 1985 at Rice University. Given the following data: CO2(g) + C(s, graphite) + O2(8) AH,° = +393.5 kJ 60 CO2 (g) → Co (s) + 60 02(8) AH2° = +25937 kJ Write the balanced standard enthalpy equation for the formation of solid Coo law to determine the standard enthalpy (or heat) formation of solid Co. Buckminsterfullere
3 attempts left Check my work Enter your answer in the provided box. Calculate AG for the reaction at 25°C. 2C6H61) + 1502(8) 12C028) + 6H20(1) agº = kJ AH° (kJ/mol Agº (kJ/mol) sº (J/K mol) 5.69 2.4 Substance C(graphite) C(diamond) CO(g) CO2(g) CO2(aq) C0,- (aq) O(g) O2(8) O3(aq) O3(8) H20(1) CH 1.90 -110.5 -393.5 -412.9 -676.3 249.4 2.87 --137.3 -394.4 -386.2 --528.1 230.1 0 16.3 163.4 -237.2 124.5 197.9 213.6 121.3 -53.1 160.95 205.0 110.88 237.6 -12.09 142.2 -285.8...