1) Enthalpy change = - 890.4 KJ
2) Enthalpy change = - 1335.6 KJ
3) Enthalpy change = - 445.2 KJ

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? Wh...
AU for the reaction CH4 (8) + 2O2(g) - CO2(8) + 2H2O(l) in kJ at 298 K is AH(CH,) =-74.6kJ / mol AH (0) = -393.5kJ / mol AH,(4,0) =-285.8kJ / mol 1 A. -885.5 kJ B.668 kJ C. 885.5 kJ D.-668 kJ
What is the enthalpy change for the first reaction? CH4(g) + 1/2O2(g) → CH3OH(g) ΔH = CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) ΔH = -802.4 CH3OH(l) + 3/2 O2 → CO2(g) + 2H2O(g) ΔH = -678.1
Use the example shown to calculate the reaction enthalpy, delta H, for the following reaction: CH4(g)+2O2(g)->CO2(g)2H2O(l). Use the series of reaction that follows: 1. C(s)+2H2(g)-> CH4(g), delta H= -74.8 kJ 2. C(s)+O2(g)->CO2(g), delta H= -393.5 kJ 3. 2H2(g)+O2(g)-> 2H2O(g), delta H= -484.0 kJ 4. H2O(l)->H2O(g), delta H= 44.0 kJ
CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔHo = -890.5 kJ. What would be the ΔHo for the reaction ½ CH4(g) + O2(g) ½ CO2(g) + H2O(l) What would be the ΔHo for the reaction ½ CH4(g) + O2(g) ½ CO2(g) + H2O(l) What would be the ΔHo for ½ CO2(g) + H2O(l) ½ CH4(g) + O2(g)
Calculate the approximate enthalpy change, ?Hrxn, for the combustion of one mole of methane a shown in the balanced chemical equation: CH4+2O2?2H2O+CO2 Use the values you calculated in Parts A, B, C, and D, keeping in mind the stoichiometric coefficients. delta H CH4=1656 kJ/mol delta H O2=498 kJ/mol delta H H2O=-928 kJ/mol delta H CO2=-1598 kJ/mol
Calculate ∆Gº for the reaction, CH4(g)+2O2(g)→CO2(g)+2H2O(g), where ∆Gfº=-50.8 kJ/mol for CH4(g), -394 kJ/mol for CO2(g), and -229 kJ/mol for H2O(g).
Calculate ∆Gº for the reaction, CH4(g)+2O2(g)→CO2(g)+2H2O(g), where ∆Gfº=-50.8 kJ/mol for CH4(g), -394 kJ/mol for CO2(g), and -229 kJ/mol for H2O(g): 572 kJ -801 kJ -572 kJ 801 kJ
When 5.00g of methan, CH4, burns 50.2 kJ of heat are produced. CH4(g)+ O2(g) > CO2() +2 H2O a. Is the reaction exothermic or endothermic? b. What is the enthalpy change of the reaction per mole of CH4? In other words, calculate the Change of enthalpy of the reaction in kJ/mol.
CH4(g) + 2O2(g), AH° = 803 kJ which of the following For the following reaction, CO2(g) + 2H2O(g) = will increase K? Select one: O a. increase the temperature of system O b. none of the above O c. decrease number of moles of methane O d. all of the above O e increase volume of system
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