Use the thermodynamic data below to answer the following questions regarding this chemical reaction: CH4(g) + H2O(g) ® 3 H2(g) + CO(g) DfG° [CH4(g)] = –50.8 kJ/mol, DfG° [H2O(g)] = –228.6 kJ/mol, DfG° [H2(g)] = 0.0 kJ/mol, and DfG° [CO(g)] = –137.2 kJ/mol; Calculate the standard net Gibbs free energy change for the reaction as written.   2. Calculate the equilibrium constant in terms of Kp and Kc. (Note: DfG°’s given above relate to Kp, not Kc. (R = 8.314 J/mol K = 0.08206 L atm/mol K)   3. Estimate the mol fraction of carbon monoxide (CO) at equilibrium?

14. Calculate the standard enthalpy change for the reaction: 2Al(s) + Fe2O3(s) 2Fe(s) + Al2O3(s) given that: 2Al(s) + 3/202(g) → Al2O3(s) 2Fe(s) + 3/202(g) → Fe2O3(s) AH°x=-1675.7kJ/mol AH x =-824.2kJ/mol

Determine the heat involved in the combustion of liquid hydrazine by using the following reactions. 2NH3(g) + 3N2O(g) → 4N2(g) + 3H2O(ℓ)      ΔH°= –1013 kJ/molN2O(g) + 3H2(g) → N2H4(ℓ) + H2O(ℓ)      ΔH°= –317 kJ/mol2NH3(g) + ½O2(g) → N2H4(ℓ) + H2O(ℓ)      ΔH°= –142.9 kJ/molH2(g) + ½ O2(g) → H2O(ℓ)       ΔH°= –285.8 kJ/mol Then, Launching a small spacecraft to study Jupiter required approximately 28 million kJ. Assume that hydrazine was the only fuel. What mass of hydrazine, in kilograms, was needed for the launch?

Given the overall reaction from the previous problem: 2 CO (g) + 6 H2 (g) D 2 CH4 (g) + 2 H2O (g) Using Le Chatelier’s principle, predict which direction the reaction will shift to reestablish equilibrium for each of the following changes: Calculate the enthalpy of reaction for this reaction using the standard thermodynamics values for the reactants and products. Is the reaction endothermic or exothermic? Now, use Le Chatelier’s principle again to determine the direction the equilibrium will shift if heat is added to the reaction.  If 35 kPa of CO is mixed with 35 kPa of hydrogen in a 1.5 L container at 25°C, what is the theoretical yield of methane?

Given the heat of formation of the following compounds:                 CO2(g)                ΔfH° = -393.5 kJ/mol                 H2O(l)                 ΔfH° = -285.9 kJ/mol                 CH3OH(l)          ΔfH° = -238.6 kJ/mol   What is the value of ΔrH° for the reaction:                 CH3OH(l) + 3/2 O2(g) → CO2(g) + 2 H2O(l)

Given the following Fe,03 (s) + 3 CO (g) →2 Fe (s) + 3 CO2 (g) AG' = - 29.4 kJ 3 Fe203 (s) + Co (g) →2 Fez04 (s) + CO, (g) AG° = -61.6 kJ Calculate AG° for Fez04 (s) + CO (g) → Fe (s) + Fe,03 (s) + CO2 (g)

Consider the reaction: 8 H2O (l) + 4CO2 (g)→4  CH3OH (l) + 6 O2 (g) ΔH = +2905.6 kJ What is ΔH for the following reaction? 2 CH3OH (l) + 3 O2 (g) →4 H2O (l) + 2 CO2 (g)

4. Calculate the enthalpy of the reaction below, given the following: CHOH 1) + 202 () → 2CO2 ®+ 3H20 ay (g) + 3H2O m AHF CO2 = -393.5 kJ/mol AHf H2O = -285.8 kJ/mol * AHf C,H5OH =-277.7 kJ/mol %3D