Hess's Law (b) Given the following thermochemical data: 2HNO3(aq) +A920(s) - H2O(1) +2AgNO3(aq) AH = -44.8...
(D) SNU29) + 12 + 2HNO3(14) + NU19) 8) Determine the overall reaction, using Hess's Law to calculate the enthalpy of the reactions from the reaction step mechanism. Step1: 300,(9) ► N269) + 02(g) AH = -99.6 kJ Step 2: H2O(l) → H2(g) + 02(9) AH; = +285.8 kJ Step 3: Hz(9) + N2(g) + 3 02(9) → 2 HNO3(aq) AH; = -414.8 kJ Step 4: N2(g) + 02(9) ► NO(9) AH; = 90.2 kJ
Hess's Law Given the following data: C2H2(g) + 5/202(9) + 2002/9) + H20(1) AH' = -1300.0 kJ CO2(g) + C(s) + O2(g) AH° = 393.5 kJ 2C(s) + H2(g) C2H2(g) AH° = 227.0 kJ Calculate AH for the reaction H2(g) + 1/202(9) — H20(1) Submit Answer Tries 0/5
The formation of nitrous acid occurs as shown below: 202(g) + N2 (g) + H2(g) 2 HNO2 (aq) Apply Hess's law and calculate the AH (in kJ) for the above reaction. Useful thermochemical data are given below. Just type the final numerical answer with proper sign and sig figs. NH4NO2 (aq) -N2(g) + 2 H20 (1) NH3(aq) + HNO2 (aq) - NH4NO2 (aq) 2NH3(aq) + N2(g) + 3 H2(g) H2(g) + 1/2O2(g) + H20 (1) AH = - 320.1 kJ...
5 Determine the standard enthalpy of formation of Fe2O3(s) given the thermochemical equations below. 2Fe(s) + 3/2 O2(g) → Fe2O3(s) AH(Fe2O3) = ? Fe(s) + 3 H2O() Fe(OH)3(s) + 3/2 H2(g) AH° = +160.9 kJ/mol-rxn H2(g) + 1/2O2(g) → H2O(1) A.H° = -285.8 kJ/mol-rxn Fe2O3(s) + 3 H2O(1) ► Fe(OH),(s) A,Hº +288.6 kJ/mol-rxn
Use Hess's law and the following data CH4(g) + 2O2(g) → CO2(g) + 2 H2O(g) AH° = -802 kJ mol-1 CH4(8) + CO2(g) —> 2CO(g) + 2 H2(g) AFH° = +247 kJ mol-1 CH4(g) + H2O(g) –> CO(g) + 3H2(g) AFH° = +206 kJ mol-1 to determine A.Hº for the following reaction, an important source of hydrogen gas CH4(8) + +02(8) — CO(g) + 2 H2(8)
DaC. NUL20 1) Find AH, for the reaction 2H2(g) + 2C(s) + O2(g) → C2H5OH(I), using the following thermochemical data. For Hess's Law, rewrite equations to find out AH. C2H5OH (1) +2 02 (g) → 2 CO2 (g) + 3H20 (1) AH = -875.J C(s) + O2(g) → CO2 (g) AH = -394,51 kJ H2(g) + 12 02 (g) → H20 (1) AH = -285.8 kJ
Please explain Data: C(graphite) + O2(g) => CO2(g) AH = -393.5 kJ H2(g) + 1/2O2(g) => H2O(1) AH = -285.8 kJ CH3OH(1) + 3/202(9) A CO2(g) + 2H20(1) AH = -726.4 kJ Using the data above, calculate the enthalpy change for the reaction below. Reaction: C(graphite) + 2H2(g) + 1/2O2(g) => CH3OH(1) A. +238.7 kJ B.-238.7 kJ C. +548.3 kJ D.-548.3 kJ E. +904.5 kJ
2. Given the following data: H2O(l) → H2(g) + 1/2O2(g) ΔH° = 285.8 kJ 2HNO3(l) → N2O5(g) + H2O(l) ΔH° = 76.6 kJ 2N2(g) + 5O2(g) → 2N2O5(g) ΔH° = 28.4 kJ Calculate ΔH° for the reaction: 1/2N2(g) + 3/2O2(g) + 1/2H2(g) → HNO3(l) Note that you should be able to answer this one without needing to use any additional information from the thermo table. I've attempted this question multiple times. I am able to get to the simplified eqaution...
6. Hess's Law. Use the data in Table 1 to calculate AH for the reaction below: Table 1. Change in Enthalpy for Reactions Reactions Change in Enthalpy (AH) AH = -393.5 kJ/mol (1) C() (2) Hz((g) + + O2(g) O2(g) → CO2(8) → H2O(l) AH2 = -285.8 kJ/mol (3) 2CH.(g) + 702(g) → 4 CO2(g)+ 6H2O(1) AH = -283.0 kJ/mol Calculate the enthalpy change for the reaction: 2 C(s) + 3H2(g) → CzH6(g) AH = kJ/mol
6. Hess's Law. Use the data in Table 1 to calculate AH for the reaction below: Table 1. Change in Enthalpy for Reactions Reactions Change in Enthalpy (AH) AH = -393.5 kJ/mol (1) C() (2) Hz((g) + + O2(g) O2(g) → CO2(8) → H2O(l) AH2 = -285.8 kJ/mol (3) 2CH.(g) + 702(g) → 4 CO2(g)+ 6H2O(1) AH = -283.0 kJ/mol Calculate the enthalpy change for the reaction: 2 C(s) + 3H2(g) → CzH6(g) AH = kJ/mol