Quizzes & Puzzles2 mins ago
How would i work this out?
4 Answers
1. Car exhaust gas catalysts are also called 3-way catalysts as they catalyse the removal
of 3 types of pollutants from the car’s exhaust gases: nitric oxides, carbon monoxide
and hydrocarbons that have not been combusted. Reaction I below shows a reaction
that may occur on such a catalyst:
Reaction I: CH4(g) + 2 NO(g) + O2(g) → N2(g) + CO2(g) + 2 H2O(g)
Use the following information to determine the enthalpy change ΔHr associated
with reaction I shown above. Note that the first enthalpy is for total combustion
of CH4(g) forming only CO2(g) and H2O(l).
ΔHc0 (CH4(g)) = - 74.81 kJ mol-1
Δ Hf0 (NO(g)) = + 90.25 kJ mol-1
ΔHvap0 (H2O(l)) = + 44.01 kJ mol-1
Authour: Hi when it says 'Note that the first enthalpy is for total combustion
of CH4(g) forming only CO2(g) and H2O(l)' Does this mean that when i work it out using ' PRODUCTS - REACTANTS, would i replace CO2 and H20 with ΔHc0 (CH4(g)) = - 74.81 kJ mol-1 as they have not given me the values for this.
of 3 types of pollutants from the car’s exhaust gases: nitric oxides, carbon monoxide
and hydrocarbons that have not been combusted. Reaction I below shows a reaction
that may occur on such a catalyst:
Reaction I: CH4(g) + 2 NO(g) + O2(g) → N2(g) + CO2(g) + 2 H2O(g)
Use the following information to determine the enthalpy change ΔHr associated
with reaction I shown above. Note that the first enthalpy is for total combustion
of CH4(g) forming only CO2(g) and H2O(l).
ΔHc0 (CH4(g)) = - 74.81 kJ mol-1
Δ Hf0 (NO(g)) = + 90.25 kJ mol-1
ΔHvap0 (H2O(l)) = + 44.01 kJ mol-1
Authour: Hi when it says 'Note that the first enthalpy is for total combustion
of CH4(g) forming only CO2(g) and H2O(l)' Does this mean that when i work it out using ' PRODUCTS - REACTANTS, would i replace CO2 and H20 with ΔHc0 (CH4(g)) = - 74.81 kJ mol-1 as they have not given me the values for this.
Answers
Best Answer
No best answer has yet been selected by HiMyNameIs. Once a best answer has been selected, it will be shown here.
For more on marking an answer as the "Best Answer", please visit our FAQ.Were the links I gave you last week on calculating enthalpy no use?
Sorry, can't help this time but I notice this thread was quite a change from the one last week when you were involved with a murder!
http://www.theanswerb...l/Question920113.html
Sorry, can't help this time but I notice this thread was quite a change from the one last week when you were involved with a murder!
http://www.theanswerb...l/Question920113.html
The link you gave me last week was useful I just asked one question which you would see if read underneath and I don't think you have any right to pass judgement. I was not involved per say I didn't not attack or murder anyone I was at the wrong place at the wrong time so if you would not mind, if your not gong to help answer THIS question, please do refrain from writing about what you no nothing about. Thank you.
-- answer removed --