Question

35. Calculate ΔrH° for the combustion of ammonia, 4 NH3(g) + 7 O2(g) → 4 NO2(g)...

35. Calculate ΔrH° for the combustion of ammonia,
4 NH3(g) + 7 O2(g) → 4 NO2(g) + 6 H2O( )
using standard molar enthalpies of formation.

molecule ΔfH° (kJ/mol-rxn)
NH3(g) –45.9
NO2(g) +33.1
H2O( ) –285.8
a. +30.24 kJ/mol-rxn
b. –206.9 kJ/mol-rxn
c. –298.6 kJ/mol-rxn
d. –1398.8 kJ/mol-rxn
e. –1663.6 kJ/mol-rxn

0 0
Add a comment Improve this question Transcribed image text
Request Professional Answer

Request Answer!

We need at least 10 more requests to produce the answer.

0 / 10 have requested this problem solution

The more requests, the faster the answer.

Request! (Login Required)


All students who have requested the answer will be notified once they are available.
Know the answer?
Add Answer to:
35. Calculate ΔrH° for the combustion of ammonia, 4 NH3(g) + 7 O2(g) → 4 NO2(g)...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Similar Homework Help Questions
  • Question 17 3 pts Using standard molar enthalpies of formation given in the table below, calculate...

    Question 17 3 pts Using standard molar enthalpies of formation given in the table below, calculate AH/xnto one decimal place, for the combustion of ammonia: AHrxn° = E nAH (products) - E mAHt"reactants) 4 NH3(g) + 7 O2(g) → 4NO2(g) + 6H2O(1) molecule AHF (kJ/mol-rxn) NH3(g) -45.9 NO2(g) +33.1 H2O(1) -285.8 H2O(9) -241.8 - 1663.6 kJ/mol-rxn +30.24 kJ/mol-rxn -1398.8 kJ/mol-rxn -298.6 kJ/mol-rxn -206.9 kJ/mol-rxn Question 11 3 pts A gas absorbs 45 kJ of heat and does 29 kJ of...

  • Please show work for the three problems... I have the answers but don't know how to...

    Please show work for the three problems... I have the answers but don't know how to do the problems. Please show each one step by step 1.) Answer: D 2.) Answer: A 3.) Answer: D Determine the heat of reaction for the combustion of ammonia 4NH3(g)+702(g)-»4NO2(g) + 6H20() using molar enthalpies of formation AH° (kJ/mol) 45.9 molecule NH3(g) NO2(g) H2O() +33.1 -285.8 a. +30.24 kJ b. -206.9 kJ c. -298.6 kJ d. -1398.8 kJ e. -1663.6 k ,Determine the heat...

  • Determine ΔrH° for the following reaction, 2 NH3(g) + 5/2 O2(g) → 2 NO(g) + 3...

    Determine ΔrH° for the following reaction, 2 NH3(g) + 5/2 O2(g) → 2 NO(g) + 3 H2O(g) given the thermochemical equations below. N2(g) + O2(g) → 2 NO(g) ΔrH° = +180.8 kJ/mol-rxn N2(g) + 3 H2(g) → 2 NH3(g) ΔrH° = –91.8 kJ/mol-rxn2 2H2(g) + O2(g) → 2 H2O(g) ΔrH° = –483.6 kJ/mol-rxn a. –1178.2 kJ/mol-rxn b. –452.8 kJ/mol-rxn c. –394.6 kJ/mol-rxn d. –211.0 kJ/mol-rxn e. +1178.2 kJ/mol-rxn

  • a) Calculate the enthalpy of the reaction 2NO(g)+O2(g)→2NO2(g) given the following reactions and enthalpies of formation:...

    a) Calculate the enthalpy of the reaction 2NO(g)+O2(g)→2NO2(g) given the following reactions and enthalpies of formation: 12N2(g)+O2(g)→NO2(g), ΔfH∘A=33.2 kJ mol−1 12N2(g)+12O2(g)→NO(g), ΔfH∘B=90.2 kJ mol−1 Express your answer with the appropriate units. b) Calculate the enthalpy of the reaction 4B(s)+3O2(g)→2B2O3(s) given the following pertinent information: B2O3(s)+3H2O(g)→3O2(g)+B2H6(g), ΔrH∘A=+2035 kJ mol−1 2B(s)+3H2(g)→B2H6(g), ΔrH∘B=+36 kJ mol−1 H2(g)+12O2(g)→H2O(l), ΔrH∘C=−285 kJ mol−1 H2O(l)→H2O(g), ΔrH∘D=+44 kJ mol−1

  • Standard Enthalpies of Formation, in kJ/mol N2(g)               0 NO2(g)            +33.2      &nbs

    Standard Enthalpies of Formation, in kJ/mol N2(g)               0 NO2(g)            +33.2                                     NH3(g) -45.9                                                   H2O(l)             -285.8 NO(g) +90.3 N2O(g)            -82.1 H2O(g)            -241.8 Use the data above to calculate ΔH for the reaction:                                 6 NO2(g) + 8 NH3(g) => 7 N2(g) + 12 H2O(g) ΔH = ?

  • The standard enthalpy change for the combustion of 1 mole of propane is -2043.0 kJ. CzH3(g)...

    The standard enthalpy change for the combustion of 1 mole of propane is -2043.0 kJ. CzH3(g) + 5 O2(g) + 3 CO2(g) + 4H2O(g) Calculate 4, Hº for propane based on the following standard molar enthalpies of formation. molecule CO2(g) H2O(g) 4,Hº (kJ/mol-rxn) -393.5 -241.8

  • For the combustion of cyclopropane at 25 ºC C3H6(g) + (9/2) O2(g) → 3 CO2(g) + 3 H2O(liq) ΔHºc(C3H6,g) = -2091 kJ / mol....

    For the combustion of cyclopropane at 25 ºC C3H6(g) + (9/2) O2(g) → 3 CO2(g) + 3 H2O(liq) ΔHºc(C3H6,g) = -2091 kJ / mol. What is the standard molar enthalpy of formation of cyclopropane(g) at 25 ºC if the standard molar enthalpies of formation of CO2(g) and H2O(liq) at that temperature are, respectively, -393.51 kJ / mol and -285.83 kJ / mol?

  • Consider the following chemical reaction and free energies of formation in the table: 4 NH3(g) +...

    Consider the following chemical reaction and free energies of formation in the table: 4 NH3(g) + 7 O2(g) → 4 NO2(g) + 6 H2O(l) Calculate ∆G°rxn (kJ) for the reaction. Only enter a numerical value.   Products or reactants ∆G°f, kJ/mol NH3(g) -17 O2(g) 0 NO2(g) +52 H2O(l) -237

  • Determine the enthalpy of formation (in kJ/mol) for NO2 (g), given the following reaction: 4 NH3...

    Determine the enthalpy of formation (in kJ/mol) for NO2 (g), given the following reaction: 4 NH3 (g) + 7 O2 (g)  4 NO2 (g) + 6 H2O(l) rxnH° = –1400 kJ, fH°(NH3, g) = –46 kJ/mol,  fH°(H2O,l) –286 kJ/mol. (1) +733 (2) +33 (3) +794 (4) –28 (5) –59

  • The combustion reaction of ethane is as follows. C2H6(g) + 7/2 O2(g) → 2 CO2(g) +...

    The combustion reaction of ethane is as follows. C2H6(g) + 7/2 O2(g) → 2 CO2(g) + 3 H2O(l) Using Hess's law and the reaction enthalpies given below, find the change in enthalpy for this reaction. reaction (1):     C(s) + O2(g) → CO2(g)     ΔH = −393.5 kJ/mol reaction (2):     H2(g) + 1/2 O2(g) → H2O(l)     ΔH = −285.8 kJ/mol reaction (3):     2 C(s) + 3 H2(g) → C2H6(g)     ΔH = −84.0 kJ/mol

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT