Calculate the Enthalpy (AH, in kJ/mol) for CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H20 (I)pased on the following known reactions (Hint: Carefully check the state symbols in your net reaction):CH4 (g) + O2 (g) → CH20 (g) + H2O (I)AH-284 kJ/molrxnСО2 (g) + H2О (9) —> CН-0 (9) + О2(9)AHrxn= + 518 kJ/molH20 (g) → H20 (I)AH– 44.0 kJ/mol= -rxn Calculate the Enthalpy (AH, in kJ/mol) for the following reaction from the heat ofrxnformation data given below (Hint: Carefully check the state symbols in this reaction):4 NO (g) + 6 H20 (I) →4 NH3 (g) + 5 O2 (g)= ?NH3(g)AH, = - 45.9 kJ/molNO(g)= + 90.3 kJ/molH20(1)AH285.8 kJ/mol= -H20(g)AH = - 241.80 kJ/molf

We’ll answer the first question since the exact one wasn’t specified. Please submit a new question specifying the one you’d like answered.Thermodynamics is the branch of chemistry that mainly deals with the heat transfer between system and surroundings. Here the system is the substance for which the heat content is studied and the remaining part of the universe is considered as the surroundings. The heat content of any substance or reaction system can be expressed in terms of enthalpy whereas the randomness in the system is associated with the entropy of the system.Hess law states that the enthalpy change for a multistep chemical reaction is the sum of all enthalpy changes. It can be written as; ∆H = H1+H2+....Given reactions: a) CH4(g) + O2(g) → CH2O(g) + H2O(l) ∆Hreaction = -284 kJ/molb) CO2(g) + H2O(g)→ CH2O(g) + O2(g) ∆Hreaction = 518 kJ/molc) H2O(g)→ H2O(l) ∆Hreaction =-44.0 kJ/molDesired reaction: CH4(g) + 2 O2(g) → CO2(g)+2 H2O(l) To calculate the enthalpy change for the reaction, reverse the equation (b): a) CH4(g) + O2(g) → CH2O(g) + H2O(l) ∆Hreaction = -284 kJ/molb) CH2O(g) + O2(g)→CO2(g) + H2O(g) ∆Hreaction = -518 kJ/molc) H2O(g)→ H2O(l) ∆Hreaction =-44.0 kJ/mol_______________________________________________________________CH4(g) + 2 O2(g) → CO2(g)+2 H2O(l) ∆Hreaction = -284-518-44=-846 kJ/mol Answer: Enthalpy change = -846 kJ/mol

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Calculate the Enthalpy (AH, in kJ/mol) for CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H20 (1) rxn ! ased on the following known reactions (Hint: Carefully check the state symbols in your net reaction): CHa (9) + O2 (g) —> CH20 (9) + H2О () AH en = - 284 kJ/mol СO: (9) + H2О (9) > CH-О (g) + O2(9) AH = + 518 kJ/mol rxn H20 (g) → H2O (1) AH = - 44.0 kJ/mol rxn

Calculate the Enthalpy (AH, in kJ/mol) for CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H20 (1) rxn ! ased on the following known reactions (Hint: Carefully check the state symbols in your net reaction): CHa (9) + O2 (g) —> CH20 (9) + H2О () AH en = - 284 kJ/mol СO: (9) + H2О (9) > CH-О (g) + O2(9) AH = + 518 kJ/mol rxn H20 (g) → H2O (1) AH = - 44.0 kJ/mol rxn

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10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

Calculate the Enthalpy (AH, in kJ/mol) for CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H20 (I)
pased on the following known reactions (Hint: Carefully check the state symbols in your net reaction):
CH4 (g) + O2 (g) → CH20 (g) + H2O (I)
AH
-284 kJ/mol
rxn
СО2 (g) + H2О (9) —> CН-0 (9) + О2(9)
AH
rxn
= + 518 kJ/mol
H20 (g) → H20 (I)
AH
– 44.0 kJ/mol
= -
rxn

Calculate the Enthalpy (AH, in kJ/mol) for the following reaction from the heat of
rxn
formation data given below (Hint: Carefully check the state symbols in this reaction):
4 NO (g) + 6 H20 (I) →
4 NH3 (g) + 5 O2 (g)
= ?
NH3(g)
AH, = - 45.9 kJ/mol
NO(g)
= + 90.3 kJ/mol
H20(1)
AH
285.8 kJ/mol
= -
H20(g)
AH = - 241.80 kJ/mol
f

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