Question

0/8 points v Previous Answers OSATOMSCHEM1 9.3.P.042. Calculate AH (in kJ/mol) for the reaction described by the equation. 4 NH3(g) +50269) + 4 NO(g) + 6 H2001 -904.0 x kJ/mol Supporting Materials Periodic Table Supplemental Data Constants and Factors Additional Materials eBook 4. -/24 points v OSATOMSCHEM1 9.3.P.068. Using the Supplemental Data, calculate the standard enthalpy change (in kJ/mol) for each of the following reactions. (a) B203(5) + 3 H2(9) ► 2B(5) + 3 H20(1) kJ/mol (b) B(s) + O2(g) → B029) kJ/mol (c) 2 NaOH(s) + CO2(g) → Na2CO3(5) + H20(9) kJ/mol (d) 4 B(s) + 3 029) 2B,03(5) kJ/mol

Answer 1

Hello, i am sorry to say that as per Chegg guidelines experts can answer only one full question that too first question only. I request you to please post remaining question as separate one since my allotted time could not permit me to answer more than one question also. Thank you in advance for your continued support to us.

Important Note: You have been not provided standard values (from your online course reference links as a table), hence i am providing exact method and how to evaluate value from trusted source (cengage) value, please double check below reference values before attempt to answer! Values are in BOLD below

Enthalpy change is denoted by ΔH symbol, so both questions 3 & 4 have same calculation method.

ΔHrxn = ∑[ΔHfo(products)] - ∑[ΔHfo(reactants)]

Given reaction: 4NH3(g) + 5O2(g) ==> 4NO(g) + 6H2O(l)

ΔHrxn = ∑[ΔHfo(4 x NO) + (6 x H2O)] - ∑[ΔHfo(4 x NH3) + (5 x O2)]

ΔHrxn = ∑[ΔHfo(4 x 90.3) + (6 x -285.8)] - ∑[ΔHfo(4 x -46.1) + (5 x 0)]

ΔHrxn = -1169.2 kJ/mol

0/8 points v Previous Answers OSATOMSCHEM1 9.3.P.042. Calculate AH (in kJ/mol) for the reaction described by the equation. 4 NH3(9) + 5 029) + 4 NO(g) + 6H20(1) kJ/mol -1169.2 - Please observe negative sign Supporting Materials Periodic Table Supplemental Data Constants and Factors Additional Materials eBook

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