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Please draw the major products and explain the mechanisms thank you
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- The enthalpy of reaction for the combustion of C to CO2 is -393.5 kJ>mol C, and the enthalpy for the combustion of CO to CO2is -283.0 kJ>mol CO:(1) C1s2 + O21g2 ---> CO21g2 ΔH = -393.5 kJ(2) CO1g2 + 12O21g2 ----> CO21g2 ΔH = -283.0 kJUsing these data, calculate the enthalpy for the combustion of C to CO:(3) C1s2 + 12 O21g2 ---->CO1g2 ΔH = ?At constant pressure and 25C, what is enthalpy for the reaction: 2C2H6 + 7O2 -> 4CO2 + H2O, if the complete consumption of 12g of C2H6 liberates - 700kJ of heat energy? (-3508 kJ)Consider the following reaction: H2(g) + ½ O2 (g) ------> H2O (g) The standard enthalpy of formation of gaseous H2O at 298 K is -241.82 kJ mol-1. Calculate the value at 153 0C. Given Cp,m for H2O(g): 33.58 kJ mol-1; H2 (g): 28.84 kJ mol-1; O2 (g): 29.37 kJ mol-1. Assume heat capacities are independent of T. NOTE:answer in kilojoules per mole (kJ/mol)
- Calculate the ΔH°f of C6H12O6(s) from the following data: Show the complete and step by step solution ΔH combustion of C6H12O6(s) = -2816kJ/mol ΔH°f of CO2 (g) = -393.5 kJ/mol ΔH°f of H2O (l) = -285.9 kJ/mol Equation: C6H12O6(s) + O2(g) -> CO2(g) + H2O(l)Calculate the enthalpy change for the reaction C(s) + 2H2 (g)--->CH4(g) from the following data. C(s) + O2 (g) ---> CO2(g). ΔH = – 393.5 kJ/mol H2(g) + 0.5 O2 (g) ---> H2O(l). ΔH = – 285.8 kJ/mol CH4(g) + 2 O2 (g)---> CO2(g) + 2 H2O(l). ΔH = – 890.2 kJ/molUsing molar enthalpies of formation, determine the heat of reaction (KJ) for the combustion of 1.0000 mole of methanol. 2CH30H() + 302(g) -› 2C02 (g) + 4H20(1) Heat f° (kJ/mol) CH3OH(1) = -238 7 C02(g) = -393 5 H20(1) = -285.8
- A pellet of benzoic acid standard was combusted in a bomb calorimeter to determine the experimental calorimeter constant. After 6 mins of monitoring the temperature, the bomb was fired. The following data were obtained. mole of benzoic acid: 0.0049 mol mass of burnt fuse wire: 0.0125 g qwire,surr: 5858 J/g ΔHcombustion of benzoic acid: -3228.0 kJ/mol Δng (combustion of benzoic acid): -0.5 mol R: 8.314 J/mol•K r1: 0ºC/min r2: 0ºC/min Tf: 30.4ºC Ti: 28.4ºC Determine the calorimeter constant (in J/ºC). Report your answer in 2 decimal places.Determine the standard heat of below reaction at 25 C. H2S(g)+2H2O(l) -> 3H2(g)+SO2(g)Calculate the change in enthalpy (in kJ/mol) for the following reaction using the data provided. 2NaHCO3(s) --> Na2CO3(s) + H2O(l) + CO2(g) ΔH°f = ? ΔH°f = NaHCO3(s) = -947.7 kJ/mol ΔH°f = Na2CO3(s) = -1131 kJ/mol ΔH°f = H2O(g) = -241.8 kJ/mol ΔH°f = H2O(l) = -285.9 kJ/mol ΔH°f = CO2(g) = -393.5 kJ/mol
- Given the following data: 4C(s) + 4H2(g) + O2(g) → CH3CH2OCOCH3(l) ΔH°=-480.0 kJ CH3CH2OH(l) + O2(g) → CH3COOH(l) + H2O(l) ΔH°=-492.0 kJ 2C(s) + 3H2(g) + 1/2O2(g) → CH3CH2OH(l) ΔH°=-278.0 kJ H2(g) + 1/2O2(g) → H2O(l) ΔH°=-286.0 kJ calculate ΔH° for the reaction:CH3COOH(l) + CH3CH2OH(l) → CH3CH2OCOCH3(l) + H2O(l)The molar heat capacity of white tin (Sn(white)) varies with temperature as follows: 20 30 40 60 80 198 249 274 289 1.970 5.460 9.387 15.33 19.49 26.17 26.71 26.84 26.88 Calculate ΔG from grey/white tin transformation at 298 K[Given: S298(Sn(grey) = 38.77 J K-1ΔH(Sn(grey) →Sn(white) = 2234 J]In constant pressure calorimeter 75.0 ml is mixed with 1.25 M Hydrocloric acid solution is mixed with 75.0 ml of 1.25 M Sodium hydroxide solution. the density of the final solution is 1.00g/ml and the the solution both initially 21.45c, reach a maximum temperature of 28.32c when mixed. based on this information, and estimating the solution's heat capacity as 4.18Jg-1c-1, what is the amount of heat in Kilojoules, transferred in this reaction?