Question

Please answer and explain!!

Once we have calculated the overall ΔHo for a reaction, we know how many k, of energy are released or absorbed as a function of the coefficients in the balanced overall equation. By relating this energy to the reaction coefficients, one can calculate the enthalpy change for a given number of grams of either reactants or products Com (in J/mol) 84.68 229.6 103.8269.9 -126.15 310.227 173.5263.47 198.78 296.018 224.4328.57 Octane (1) (CgH18 (molecular formula), CH3CH2CH2CH2CH2CH2CH2CH3 (organic structure))249.95 361.205 274.7 393.67 mol Ethane (g) (C2H6 (molecular formula), CH3CH3 (organic structure)) Propane (9) (C3Hg (molecular formula), CH3CH2CH3 (organic structure)) Butane (9) (C4H10 (molecular formula), CH3CH2CH2CH3 (organic structure)) Pentane (I) (CSH12 (molecular formula), CH3CH2CH2CH2CH3 (organic structure)) Hexane (1) (C6H14 (molecular formula), CHJCH2CH2CH2CH2CH3 (organic structure)) Heptane (I) (C H16 (molecular formula), CH3CH2CH2CH2CH2CH2CH3 (organic structure)) Nonane (I) (CgH20 (molecular formula), CH3CH2CH2CH2CH2CH2CH2CH2CH3 (organic structure)) Carbon dioxide (g) (CO2) Water (g) (H20) 393.509 213.74 241.818 188.825 a) using the 솨to values in the table provided, calculate the ΔHo for the combustion of liquid C7H16 (heptane) to produce gaseous c02 and H20 at standard state conditions? Balance your combustion equation using whole number coefficients 2890.6 Do you wish to review this material from the appropriate tutorial? b) what is the 솨 if 3.80 grams of liquid C7H16 (heptane) are completely combusted? Not sure how to start? Go back and read the small paragraph at the very beginning of this problem. 130.8

Answer 1

Ans. #A. Balance reaction:   C₇H₁₆(l) + 11 O₂(g) ---->7 CO₂(g) + 8 H₂O(g)

dH⁰ combustion = Sum of dH_f⁰ of products - Sum of dH_f⁰ of reactants

Or, dH⁰= (7 mol x dH_f⁰ of CO₂ + 8 mol x dH_f⁰ of H₂O) – (dH_f⁰ C₇H₁₆ + 11 mol x dH_f⁰ of O₂ )

Or, dH⁰ = [ 7 mol x (-393.509 kJ mol^-1) + 8 mol x (-241.818 kJ mol^-1)] –

[ ( - 224.4 kJ mol^-1) + 11 mol x 0 )

Or, dH⁰ = -4689.107 kJ mol^-1 + 224.4 kJ mol^-1 = -4464.707 kJ mol^-1

Hence, dH⁰ of combustion of heptane = -4464.707 kJ mol^-1

#B. Given, mass of heptane = 3.80 g

So,

            Moles of heptane = Mass / Molar mass

                                                = 3.80 g / (100.20404 g/mol)

                                                = 0.037923 mol

# Now, dH⁰ of combustion of 3.80 g heptane =

Moles of heptane x molar enthalpy of combustion of heptane

= 0.037923 mol x (- 4464.707 kJ mol^-1)

= - 169.31 kJ