The standard heat of reaction for the combustion of n-butane vapor is 2C4H10(g)

Question

The standard heat of reaction for the combustion of n-butane vapor is 2C4H10(g) + 13O2(g) ----> 8CO2(g) + 10H2O(l) delta Hr=-5756kJ/mol If the reactants enter at 25

Explanation / Answer

-5492 kJ/mol. 2C4H10(g) -> (delta)H?f = -252.3 kJ/mol , S?m = 620.2 J/mol·K 13O2(g) -> (delta)H?f =0 , S?m = 2666.82 J/mol·K 8CO2(g) -> (delta)H?f = -3148.08 kJ/mol , S?m = 1709.92 J/mol·K 10H2O(l) -> (delta)H?f = -2858.3 kJ/mol , S?m = 699.1 J/mol·K T?S 2C4H10(g) -> T*S?m = 298.15 K*620.2 J/mol·K = 184912.63 J/mol 13O2(g) -> T*S?m = 298.15 K*2666.82 J/mol·K = 795112.383 J/mol 8CO2(g) -> T*S?m = 298.15 K*1709.92 J/mol·K = 509812.648 J/mol 10H2O(l) -> T*S?m = 298.15 K*699.1 J/mol·K = 208436.665 J/mol G = H-TS 2C4H10(g) -> ?G?r1 = H - T?S = -252300 - 184912.63 = -437212.63 13O2(g) -> ?G?r2 = H - T?S = 0 - 795112.383 = -795112.383 8CO2(g) -> ?G?r3 = H - T?S = -3148080 - 509812.648 = -3657892.648 10H2O(l) -> ?G?r4 = H - T?S =-2858300 - 208436.665 = -3066736.665 ?G?r = (?G?r4 + ?G?r3) - (?G?r2+?G?r1) = -3066736.665-3657892.648+795112.383+4372… ?G?r = - 5492304.3 J/mol = -5492.3043 kJ/mol

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