3. For a particular isomer of C8H18, the following reaction produces 5104.lkJ of
ID: 855229 • Letter: 3
Question
3. For a particular isomer of C8H18, the following reaction produces 5104.lkJ of heat per mole of C8H18(g) consumed, under standard conditions (equation shown below). What is the standard enthalpy of formation of this isomer of C8H18(g) (in kJ/mol)? You may use the standard enthalpies of formation given in the table above. C8H18(g) + 25/2 O2(g) --> 8 CO2(g) + 9 H2O(g) deltaH degree rxn = -5104.lkJ 4. Given that: H2(g) + F2(g) --> 2 HF(g) deltaH degree rxn = -546.6kJ 2 H2(g) + 02(g) --> 2 H2O(l) deltaH degree rxn = -571.6kJ Calculate the value of deltaH degree rxn (in kJ) for the following reaction: 2 F2(g) + 2 H2O(l) --> 4 HF(g) + O2(g)Explanation / Answer
1.
Standard for of equation is to obtain the compound from the elements in their standard states.
Na(s) + 1/2N2(g) + 3/2O2(g) ---------> NaNO3(s)
2.
Delta H = Delta H(products) - Delta H(reactants)
= Delta H(CO2) - [(Delta H CO) +(Delta H H2O)]
=-393.5 - [(-110.5-241.8)]
= -41.2 kJ/mol
This is for 1 mol of CO (28 g CO)
Delta H in kJ/mol = -41.2 / 28 = -1.471 kJ
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3.
Delta H for C8H18 = -5104.1kJ
molar mass of C8H18 is 114 g/mol
Delta H of C8H18 in kJ/mol = -5104.1 / 114 = -44.77 kJ/mol
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4.
H2(g) + F2(g) -----> 2HF(g) delta H = -546.6 kJ ---------eq(1)
2H2(g) + O2(g) -------> 2H2O(l) delta H = -571.6 kJ ----------eq(2)
eq(1) multiplied by2
eq(2) is reversed
now. add them.
2H2(g) + 2F2(g) -----> 4HF(g) delta H = 2 x -546.6 kJ
2H2O(l) -------> 2H2(g) + O2(g) delta H = +571.6 kJ
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2F2(g) + 2H2O(l) -------> 4HF(g) + O2(g) Delta H = -521.6 kJ
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Delta H = (2 x -546.6) + (+571.6)
= -521.6 kJ
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