Question

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12. What is of formations provided? nolar enthalpy change AHnu for the following reactions using the standard molar enthalpies sich() + 2 11,0 (l) ? SiO2 (s) + 4 HCl (g) AH (SiCIA) -640.1 kJ/mol AH (SiO2)-910.9 kJ/mol AH (H2O) -285.8 kJ/mol AH (HCI)-92.3 kJ/mol a. -2491.8 kJ/mol b. -1929.1 kJ/mol c. -77.3 kJ/mol d. -68.4 kJ/mol e. None of these 13. Given the following thermochemical equations and their corresponding enthalpies: 2 C H2 (g) +5 02(g) 4 CO2 (g) +2 H,o() CsHs (t)+10 02 (g)-8 CO2 (8)+4 H2O () AH -2601 kJ ??_-4393 kJ What is the enthalpy change for the following reaction? 4 C2H2 (g)-C8Hs () AH-? a. -1792 kJ b. -809 kJ C. -6994 kJ d. -9595 k.J e. none of these 14. Evaluate Af for the following reaction from the given bond energies. 2HBr(g) ->H2(g)+ Br2g) AHH-H-436 kJ/mol, Afr-Br-193 kJ/mol, AHH-Br-366 kJ/mol a. -103 k.J b. -143 kJ c. +103 kJ d. +142 kJ e. 259 kJ

Answer 1

1) SiCl₄ + 2H₂O -------> SiO₂ + 4HCl

$\Delta$ H_f(SiCl4) = -640.1kJ/mol $\Delta$ H_f(H2O) = -258.8 kJ/mol $\Delta$ H_f(SiO2) = -910.9kJ/mol   $\Delta$ H_f(HCl) = -92.3kJ/mol

$\Delta$ H_rxn = $\sum$ n $\Delta$ H_f(products) - $\sum$ n $\Delta$ H_f(reactants)

$\sum$ n $\Delta$ H_f(products) = -910.9+ 4(-92.3) = -1280.1 kJ/mol

$\sum$ n $\Delta$ H_f(reactants) = -640.1+ 2(-285.8) = -1211.7 kJ/mol

$\Delta$ H_rxn = -1280.1 - (-1211.7) = -68.4 kJ/mol

2) 2C₂H₂ + 5O₂ -----------> 4CO₂ + 2H₂O    $\Delta$ H = -2601 kJ ------- MULTIPLY BY 2

C₈H₈ + 10O₂ -----------> 8CO₂ + 4H₂O    $\Delta$ H = - 4393 kJ ------ REVERSE THE REACTION

2( 2C₂H₂ + 5O₂ -----------> 4CO₂ + 2H₂O)    $\Delta$ H = 2(-2601) kJ

4C₂H₂ + 10O₂ -----------> 8CO₂ + 4H₂O     $\Delta$ H = -5202 kJ ---- 1

8CO₂ + 4H₂O -----------> C₈H₈ + 10O₂ $\Delta$ H = +4393 kJ ----- 2

Adding equation 1 and 2 gives the equation 3

   4C₂H_2(g) ------> C₈H_8(l) --------------- 3   $\Delta$ H = ? kJ

$\Delta$ H = $\Delta$ H_eq1 + $\Delta$ H_eq2 = -5202+4393 = -809 kJ

3) 2HBr -----> H₂ + Br₂

$\Delta$ H_f(H-H) = 436 kJ/mol $\Delta$ H_f(Br-Br) = 193 kJ/mol $\Delta$ H_f(H-Br) = 366 kJ/mol

$\Delta$ H_rxn = $\sum$ n $\Delta$ H_f(products) - $\sum$ n $\Delta$ H_f(reactants)

$\sum$ n $\Delta$ H_f(products) = 436 + 193  = 629 kJ/mol

$\sum$ n $\Delta$ H_f(reactants) = 2(366) = 732 kJ/mol

$\Delta$ H_rxn = 732 - 629 = +103 kJ/mol