Solid benzoic acid, C6H5COOH(s), is often used to determine the heat capacity of

Question

Solid benzoic acid, C6H5COOH(s), is often used to determine the heat capacity of a bomb calorimeter, because the constant volume heat of combustion of this compound is known with high precision:
qV = 26.434 kJ/g. When a 2.088 g sample of C6H5COOH(s) was burned in a particular bomb calorimeter, the temperature increased by 11.77 oC. In a second experiment, a 1.217 g sample of a compound with the formula C6H14O(l) was burned in the same calorimeter. The temperature increased by 10.08 oC. Use these data, plus the information given below for CO2 and H2O, to estimate the standard enthalpy of formation, of H, at 25 oC for this compound. of H values at 25 oC for CO2(g) and H2O(l) are 394 kJ mol-1 and 286 kJ mol-1,

Explanation / Answer

heat released when 2.088 gm of benzoic acid was burned = 2.088*26.434 = 55.194192 KJ

so heat capacity of calorimeter = 55.194192/temp change
= 55.194192/11.77 = 4.69 KJ/K

so when temp of calorimeter rises by 10.08 K,
so heat released = 4.69*10.08 = 47.27 KJ

molar mass of C6H14O = 102 gm/mol
so moles of C6H14O = 1.217/102 = 0.012 mol

so delta H(rxn) = 47.27/0.012 = -3939.17 KJ/mol

also,
C6H14O + 9O2 ---------> 6CO2 + 7H2O

so, delta H(rxn) = 6(-394)+7(-286)-x ...(where x is delta H formation of C6H14O)

so,
-3939.17 = -4366-x
solving, x = -426.83 KJ/mol ...... is the answer

you might have got -404.25 KJ/mol due to some rounding errors, but as per above calculation nothing seems wrong, the answer should be -426.83 KJ/mol

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