Calculate the standard enthalpy of formation of NO(g) (in kJ/mol) from the following data.
N2(g) + 2 O2(g) → 2 NO2(g)
ΔH⁰298
= 66.4 kJ 2 NO(g) + O2(g) → 2 NO2(g)
ΔH⁰298
= −116.2 kJ
On reversing the equation,dH gets changed the sign. That means if forward reaction is endothermic then backward reaction is exothermic.
Calculate the standard enthalpy of formation of NO(g) (in kJ/mol) from the following data. N2(g) +...
The standard enthalpy change for the following reaction is 66.4 kJ at 298 K. N2(g) + 2 O2(g) 2 NO2(g) AH° = 66.4 kJ What is the standard enthalpy change for this reaction at 298 K? 1/2 N2(g) + O2(g) — NO2(g) Submit Answer
A.Using standard heats of formation, calculate the standard enthalpy change for the following reaction. N2(g) + 3H2(g) = 2NH3(g) B.Using standard heats of formation, calculate the standard enthalpy change for the following reaction. CaCO3(s) = CaO(s) + CO2(g) C. A scientist measures the standard enthalpy change for the following reaction to be -2910.6 kJ: 2C2H6(g) + 7 O2(g) = 4CO2(g) + 6 H2O(g) Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy...
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 = ?
4. The standard enthalpy of formation of NH3 (g) is -46.11 kJ mol-' at 298 K. Given the heat capacity data below and the data in Problem 2, calculate the standard enthalpy of formation at 1200 K Cp.m (H2 (9))/ J mol K-1 = 29.1 - (0.84 x 10- K-)T Cpm (N2 (g))/ J mol K-1 = 26.98 +(5.9 x 10-'K-!)T
The standard enthalpy change for the following reaction is -181 kJ at 298 K. 2 NO(g) -----------> N2(g) + O2(g) ΔH° = -181 kJ What is the standard enthalpy change for this reaction at 298 K? 1/2 N2(g) + 1/2 O2(g) ----------> NO(g)
Calculate the enthalpy of the reaction 2NO(g)+O2(g)→2NO2(g) Given the following reactions and enthalpies of formation: N2(g) + 2O2(g)→ 2NO2(g), ΔH∘ = 66.4kJ N2(g)+ O2(g)→ 2 NO(g), ΔH∘=180.4 kJ Please explain the steps as well!
The standard enthalpy of formation of CO(g) is −111 kJ/mol [refer to this equation C(s) + 1/2O2(g) ⟶ CO(g)] What is the ΔHorxn of this reaction? (What is the standard enthalpy change of this reaction?) 2C(s) + O2(g) ⟶ 2CO(g) -253 kJ/mol - 222 kJ/mol -97.1 kJ/mol 534 kJ/mol 124.9 kJ/mol -50.5 kJ/mol 5 points QUESTION 2 (assume A, B, C, and D and E are chemicals) Look at these equations A + B à C ΔH° = -9 kJ/mol rxn...
Use the following data to calculate the standard enthalpy of formation of heptane, C7H16 (l). C7H16 (l) + 11 O2 (g → 7 CO2 (g) + 8 H2O (l) ΔH° = -4817 kJ/mol ΔHf° of CO2 (g) = -393.5 kJ/mol ΔHf° of H2O (l) = -285.8 kJ/mol A)-218.2 kJ/mol B)-468.1 kJ/mol C)-223.9 kJ/mol D)-447.8 kJ/mol E)-111.5 kJ/mol
c. Calculate ΔH° for the process Co3O4(s)→3Co(s) + 2O2(g) from the following information: Co(s) + 1/2O2(g)→CoO(s) ΔH° = -237.9kJ 3CoO(s) +1/2O2(g)→Co3O4(s) ΔH° = -177.5k d. Calculate the standard molar enthalpy of formation of NO(g) from the following data: N2(g) + 2O2 →2NO2(g) ΔH° = 66.4 k. 2NO(g) + O2 →2NO2(g) ΔH° = -114.1 kJ
Consider the following chemical reaction. NH3(g) + 2 O2(g) → HNO3(aq) + H2O(l) Calculate the change in enthalpy (ΔH) for this reaction, using Hess' law and the enthalpy changes for the reactions given below. (1a) 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(l); ΔH = −1166.0 kJ/mol (2a) 2 NO(g) + O2(g) → 2 NO2(g); ΔH = −116.2 kJ/mol (3a) 3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g); ΔH = −137.3 kJ/mol