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Use the combustion of hydrogen with oxygen to form water to answer questions 19 and 20. H_2(g) + 1/2 O_2(g) rightarrow H_2O(l) What is the standard change in free energy (Delta G degree) and spontaneity at 25 degree C for the reaction above? A. -237.2 kJ: spontaneous B. -237.2 kJ: not spontaneous C. 237.2 kJ: spontaneous D. 237.2 kJ: not spontaneous What will happen at elevated temperatures? A. The reaction will be non-spontaneous because Delta S > 0. B. The reaction will be non-spontaneous because Delta S 0. D. The reaction will be spontaneous because Delta S

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Casey DurganLv2

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Related questions

Given the following thermodynamic data and balanced equation, calculate the standard free energy change (delta G degree) for the reaction at 25 degree C. Also, based on the calculated (delta G degree), is the reaction spontaneous or non- spontaneous at this temperature? [delta G degree = delta H degree - (delta S degree middot T)] CH_4 (g) + 2 O_2 (g) rightarrow CI_2 (g) + 2 H_2 O(I) delta H degree = -806 kJ {delta S degree [CO_2 (g) = 213.6 J/mol middot K]}, {delta S degree [H_2 O(I) = 129.3 J/mol middot K}}, {delta S degree [CH_4 (g) = 186.2 J/mol middot K]}, {delta S degree [O_2 (g) = 205 J/mol middot K]} a) 0.0 kJ: Reaction is at Equilibrium b) 268.6 kJ: Reaction is Non-Spontaneous c) -268.6 kJ: Reaction is Spontaneous d) -806 kJ: Reaction is Spontaneous e) 806 kJ: Reaction is Non-Spontaneous Given the following thermodynamic data and balanced equation, calculate the standard free energy change (delta G degree) for the reaction at 25 degree C. Also, based on the calculated (delta G degree), is the reaction spontaneous or non-spontaneous at this temperature? [delta G degree = delta H degree - (delta S degree middot T)] 2 Mg O(s) rightarrow 2 Mg (s) + O_2 (g) delta H degree = 1203 kJ {delta S degree [Mg (s) = 32.5 J/mol middot K] }, {delta S degreeÂÂ [O_2 (g) = 205 J/mol middot K]}, {delta S degree [MgO(s) = 26.8 J/mol middot K]} a) 1139 kJ: Reaction is Non-Spontaneous b) -1139 kJ: Reaction is Spontaneous c) 216.4 kJ: Reaction is Non-Spontaneous d) -216.4 kJ: Reaction is Spontaneous e) 472 kJ: Reaction is Non-Spontaneous Ozone in the lower atmosphere is a pollutant that can form by the following reaction involving the oxidation of unburned hydrocarbons. Using the following thermodynamic data at 25 degree C, calculate the standard change in Free Energy (delta G degree) for the following reaction: CH_4(g) + 8O_2(g) rightarrow ÂCO_2(g) + 2H_2 O(g) + 4O_3(g) delta G degree = ? {delta G degree [CO_2 (g) = 394.4 kJ/mol]}, (delta G degree [H_2 O(g) = -228.6 kJ/mol]}, {delta G degree [O_3(g) = 163.2 kJ/mol]}, {delta G degree ÂÂ[CH_4 (g) =-50.5 kJ/mol]}, {delta G degree [O_2(g) = 0.0 kJ/mol]} a) -465.0 kJ b) 148.4 kJ c) -114.9 kJ d) 114.9 kJ e) -148.4 kJ

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