please explain each step in clear writing.
please explain each step in clear writing. Chloroform (CHCI3) is formed from methane and chlorine in...
717)Use the reactions below to calculate the standard change in enthalpy for the reaction of methane (CH4) with chlorine to form chloroform (CHC13) and hydrochloric acid: CH4(g) + 3Cl2(g) → CHCI3(g) + 3HCI(g) Reaction 1: HCI(g) → H2(g) + 2Cl2(g) AH° = +92.3 kJ Reaction 2: CH4(g) → C(s) + 2H2(g) AH° = +74.8 kJ Reaction 3: C(s) + 3H2(g) + 3/2Cl2(g) →CHCl3(g) AH° = -103.1 kJ x a. +270.2 kl b. +248.6 kJ PC. +64.0 kJ -120.6 kJ -305.2...
Calculate the standard enthalpy of formation of gaseous methane (CH4) using the following thermochemical information: CO2(g) + 2 H2O(l) CH4(g) + 2 O2(g) H = +890.4 kJ CO2(g) C(s) + O2(g) H = +393.5 kJ 2 H2O(l) 2 H2(g) + O2(g) H = +571.7 kJ
From the following data, C(graphite) + O2(g) → CO2(g) AHrxn = -393.5 kJ/mol H2(g) + O2(g) → H200) AH"rxn = -285.8 kJ/mol 2C2H6(g) + 702(g) -> 4CO2(g) + 6H2O(l) Arxn=-3119.6 kJ/mol Calculate the enthalpy change for the reaction: 2 C(graphite) + 3H2(g) + C2H668)
CH_(g) + 2O2(g) → CO2(g)+ 2H2O(1) AH = -890.4 kJ When 1 mole of CO2(g) is produced, what is the enthalpy change? When 3 mol of O, is reacted with an excess of CH4, what is the enthalpy change? When 1 mole of H,O(l) is produced, what is the enthalpy change?
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()
From the following heats of combustion, CH3OH(l) + 3/2O2(g) → CO2(g) + 2H2O(l) ΔHorxn = –726.4 kJ/mol C(graphite) + O2(g) → CO2(g) ΔHorxn = –393.5 kJ/mol H2(g) + ½O2(g) → H2O(l) ΔHorxn = –285.8 kJ/mol Calculate the enthalpy of formation of methanol (CH3OH) from its elements. C(graphite) + 2H2(g) + ½O2(g) → CH3OH(l) Methanol (CH3OH) is an organic solvent and is also used as a fuel in some automobile engines. From...
Please explain Data: C(graphite) + O2(g) => CO2(g) AH = -393.5 kJ H2(g) + 1/2O2(g) => H2O(1) AH = -285.8 kJ CH3OH(1) + 3/202(9) A CO2(g) + 2H20(1) AH = -726.4 kJ Using the data above, calculate the enthalpy change for the reaction below. Reaction: C(graphite) + 2H2(g) + 1/2O2(g) => CH3OH(1) A. +238.7 kJ B.-238.7 kJ C. +548.3 kJ D.-548.3 kJ E. +904.5 kJ
Heat of Formation Calculations: 32) Use a standard enthalpies of formation (Ho) table to determine the change in enthalpy for each of these reactions Hrxn [n. Ho(products) - n. Ho(products)] CO (g): -110.5 kJ/mol; CO2 (g): -393.5 kJ/mol CH4 (g): -890.4 kJ/mol H2O (l): -285.8 kJ/mol; H2O (g): -241.8 kJ/mol H2S (g): -20.6 kJ/mol; NO: -90.2 kJ/mol NO2: +33.9 kJ/mol; HCl (g): -92.3 kJ/mol NaOH (s): -426.7 kJ/mol; SO2 (g): -296.8 kJ/mol a) CH4(g) + 2 O2(g) ---> CO2(g) +...
The overall reaction to produce chloroform from methane is CH4(g) + 3Cl2(g) ---> CHCl3(l) + 3HCl(g). Use the following intermediate reactions to calculate the enthalpy of the reaction, Hrxn. 1/2 H2(g) + 1/2 Cl2(g) ---> HCl(g) Hrxn = -92.3 kj C(s) + 2H2(g) ----> CH4(g) Hrxn = -74.8 kj C(s) + 1/2 H2(g) + 3/2 Cl2(g) ----> CHCl3(l) Hrxn = -134.5 kj
Use the Data table to calculate ∆H for the reaction below:Reactions: Change in Enthalpy (∆H)(1) C (s) + O2 (g) -> CO2(g) ∆H1 = -393.5 kJ/ mol(2) H2 (g) + 1/2 O2 (g) -> H2O (l) ∆H2 = -285.8 kJ/mol(3) 2C2H6 (g) + 7O2 (g) -> 4 CO2 (g) + 6 H2O (l) ∆H3 = -283.0 kJ/molCalculate the enthalpy change for the reaction:2 C (s) + 3 H2 (g) -> C2H6(g) ∆H = ______________kJ/mol