Acetone (C_3H_6O (I)) burns in oxygen according to the equation: 4 O_2(g) + C_3H

Question

Acetone (C_3H_6O (I)) burns in oxygen according to the equation: 4 O_2(g) + C_3H_6O(I) = 3 CO_2(g) + 3H_2O(I) At 25 degreeC the enthalpy change for this reaction is -1789.9 kJ. The standard enthalpy of formation Delta_fHdegree of CO_2 (g) and H_2O (I) are -393.5 1 and -285.83 kJ/mol, respectively. Determine the enthalpy of formation of C_3H_6O(I) at 25 degreeC. I really need help on answering the continued ased on the answer from above. CONTINUED.. Using the result from last question to calculate the reaction enthalpy Delta_rH at 0degreeC (instead of 25degreeC, as given in the last question), if the heat capacities at constant pressure (C_p) of C_3H_6O(I), O_2(g), CO_2(g), and H_2O(I) are 124.7J K^-1 mol1^-1, 29.36J K^-1 mol^-1, 29.10J K^-1 mol^-1, and 75.29J K^1 mol^-1, respectively. Please enter your answer with 2 decimals as 23.22. Using KJ/mol as the unit. Pay attention to the sign.

Explanation / Answer

1. Standard enthakpy change for a reaction,

dHrxn = dHproducts - dHreactants

-1789.9 = (3 x -393.51 + 3 x -285.83) - (dHof[C3H6O])

dHof[C3H6O] = (3 x -393.51 + 3 x -285.83) + 1789.9 = -248.12 kJ

dT = T2 - T1 = 298 - 273 = 25 K

dCprxn = [dCpproducts - dCpreactants]        

             = (3 x 29.36 + 3 x 75.29] - [124.7 + 4 x 29.36]

             = 71.81 J/K.mol

dCprxndT = 0.07181 x 25 = 1.79525 kJ/mol

So, enthalpy change at 0 oC would be,

dH[0 oC] = -248.12 + 1.79525 = -246.325 kJ

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