Unformatted text preview: points)
ΔS° = ΔS°products -‐ ΔS°reactants = 2*(188.7) – [2*(130.6) + (205)] = -‐89 J/mol·∙K (10 points, -‐5 per major mistake i.e. stoichiometry, order, math I could not follow) Yes, 3 moles of gas converting into 2 moles of gas is an overall decrease in possible degrees of freedom, so entropy is negative for the system. c) H2 and O2 are at their reference states, so what are their ΔH° values?
Given ΔH°(H2O) = –241.8 kJ/mol, what is ΔH° for the reaction? Based on your answer, does this process absorb or release heat? (circle one) (5 points)
ΔH° = ΔH°products -‐ ΔH°reactants = 2*(-‐242) – [2*(0) + (0)]= -‐484 kJ/mol (10 points, -‐5 per major mistake i.e. stoichiometry, incorrect ΔH for O2/N2, order, math I could not follow) This reaction releases a large amount of heat. d) Calculate ΔG° for the reaction. Is this reaction spontaneous at 298 K? Yes or No (5 points)
ΔG° = ΔG°products -‐ ΔG°reactants ΔG° = ΔH° -‐ T·∙ΔS° = -‐484 kJ/mol – (298K*-‐0.089kJ/mol·∙K)...
View Full Document