CH4(g) + 2O2(g), AH° = 803 kJ which of the following For the following reaction, CO2(g)...
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
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).
For the reaction CH4 + 2O2 → CO2 + 2H2O, how many moles of carbon dioxide are produced from the combustion of 97.0 g of methane?
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
Burning 1.00 mol of methane releases 803 kJ of energy. How much energy is released by burning 27.7 mol of methane? CH4(g)+2O2(g)→CO2(g)+2H2O(g)+803kJ Express your answer with the appropriate units.
Methane, CH4, reacts with oxygen to produce carbon dioxide, water, and heat. CH49) + 2O2(g) - CO2(g) + 2H2O() What is the value of AH if 5.00 g of CH4 is combusted? 157 kJ 277 kJ 445 kJ -714 kJ 1.43 104 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? 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?
The thermochemical equation of combustion of methane is: CH4(g) + 2O2(g) → CO2(g) + 2 H2O(l) ΔΗ =-890.3 kJ 1. Calculate the AH when 5.00 g CH4 react with excess of oxygen. 2. Calculate AH when 2L CH4 at 49 °C and 782 mmHg react with an excess of oxygen 3. Calculate AH when 2L CH4 react with L O2 in a reaction vessel kept at 49 °C and 782 mmHg.
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
Use Hess's law and the following data CH4(g) + 2O2(g) → CO2(g) + 2 H2O(g) AH° = -802 kJ mol-1 CH4(8) + CO2(g) —> 2CO(g) + 2 H2(g) AFH° = +247 kJ mol-1 CH4(g) + H2O(g) –> CO(g) + 3H2(g) AFH° = +206 kJ mol-1 to determine A.Hº for the following reaction, an important source of hydrogen gas CH4(8) + +02(8) — CO(g) + 2 H2(8)