Question

Example 17: Fill in the chart: 101 Version A Spring 2019 This Problem continues for twe.pages Natural gas ( mostly CH.) accounts for 48% of the world's industrial hydrogen production. Hydrogen can be generated from natural as a 80efficiency. Hydrogen is often produced by the steam reforming of methane. This process consists of heating the gas in the presence of steam and a nickel catalyst Step A: CH. (g) + H2O(g) → CO(g) + 3 H2(g) (reaction at 1700K) The carbon monoxide gascan then be passed with steam over iron oxide and undergo a water gas shin reaction. Step B: CO (8)+ H2O(g) → CO2(g) + H2(g) This last reaction produces more H. The downside to this process is that its major byproducts are CO2 and other greenhouse gases. Depending on the quality of the natural gas, one ton of hydrogen produced may produce 9 to 12 tons of CO2. company hires you to design an industrial facility to produce fuel in two steps above... Thermodynamike wawesi swbstance AN AJ/mol AG J/mol Si J/mol K -74.6 50.5 186.3 H2O(g) -241.8 -228.6 188.8 CO (8) -110.5 -137.2 197.7 H2(g) Intentionally blank Intentionally blank 130.7 CO2 (g) -393.5 -394.4 213.8 CH. (1 pt)a) What are AH®, and AG for H, gas in standard conditions? (agreed on by scientists) For H AHO AGⓇO b) For Step A: CH. (g) + H:0 (g) → CO(g) + 3 H2(g) (2 pt) Solve for enthalpy change of reaction A, AHA 42.9 Mathon AHA-1360) + (-110.5)]--44) (-2415)] (2 pt) Solve for entropy change of reaction A, AS, 214.7 (Check to see if the sign of ASørxn fits what you expect for the reaction) ASA: 3 (130.9) +1939) - [15.+ 118.8 ] (3 pt) Solve for an equilibrium temperature if there is one and specifically circle only one of the conditions for step A to be spontaneous? a)always, flower than equilibrium T, c)higher than equilibrium T. d) never if there is one, the equilibrium temperature is ...Te = (2 pt) c) Solve for Gibb's free energy for step A, AGA at 25°C_1419 (310)+(-137.2 fl-50.5+ (-228. 6 2 99.1 Is this spontaneous under standard conditions? Yes or No (at 25°C) no

Please help with #17!!!!

Answer 1

1.b) $\Delta$H_A^o = -110.5 - (-316.4) = +205.9 kJ/mol

$\Delta$S_A^o = +214.7 J/mol.K

The reaction is spontaneous at T > ($\Delta$​​​​​​H/$\Delta$S)

i.e. T > T_e or T_e < T

1.c) The reaction is spontaneous at T > 205.9/0.2147, i.e. T > 959 K

1.d) $\Delta$G = $\Delta$H - T.$\Delta$S

i.e. $\Delta$G_A^o at 1100 K = 205.9 kJ/mol - 1100 K * 0.2147 kJ/mol.K = -30.27 kJ/mol (negative)

Hence, the reaction is spontaneous at 1100 K.

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