Question

Using values from Appendix C in the textbook, calculate the standard enthalpy change for each of the following reactions.

Link to Appendix C: https://media.pearsoncmg.com/ph/esm/esm_brown_chemistry_14/appendices/appendix-c.pdf

1. 2SO2(g)+O2(g)→2SO3(g)

2. Mg(OH)2(s)→MgO(s)+H2O(l)

3. N2O4(g)+4H2(g)→N2(g)+4H2O(g)N2O4(g)+4H2(g)→N2(g)+4H2O(g)

4. SiCl4(l)+2H2O(l)→SiO2(s)+4HCl(g)

Answer 1

(1.) Balanced equation : 2 SO₂ (g) + O₂ (g) $\rightarrow$ 2 SO₃ (g)

Standard enthalpy change, $\Delta$ H^o = $\Delta$ H^o_f products - $\Delta$ H^o_f reactants

$\Delta$H^o = 2 * $\Delta$ H^o_f SO₃ (g) - [2 * $\Delta$ H^o_f SO₂ (g) + $\Delta$ H^o_f O₂ (g)]

$\Delta$H^o = 2 * (-395.2 kJ/mol) - [2 * (-296.9 kJ/mol) + (0)]

$\Delta$H^o = -790.4 kJ + 593.8 kJ

$\Delta$H^o = -196.6 kJ

(2.) Balanced equation : Mg(OH)₂ (s) $\rightarrow$ MgO (s) + H₂O (l)

Standard enthalpy change, $\Delta$ H^o = $\Delta$ H^o_f products - $\Delta$ H^o_f reactants

$\Delta$H^o = [$\Delta$H^o_f MgO (s) + $\Delta$ H^o_f H₂O (l)] - $\Delta$ H^o_f Mg(OH)₂ (s)

$\Delta$H^o = [(-601.8 kJ/mol) + (-285.83 kJ/mol)] - (-924.7 kJ/mol)

$\Delta$H^o = -887.63 kJ + 924.7 kJ

$\Delta$H^o = 37.07 kJ

(3.) Balanced equation : N₂O₄(g) + 4 H₂(g) $\rightarrow$ N₂(g) + 4 H₂O(g)

Standard enthalpy change, $\Delta$ H^o = $\Delta$ H^o_f products - $\Delta$ H^o_f reactants

$\Delta$H^o = [$\Delta$H^o_f N₂(g) + 4 * $\Delta$ H^o_f H₂O(g)] - [$\Delta$H^o_f N₂O₄(g) + 4 * $\Delta$ H^o_f H₂(g)]

$\Delta$H^o = [(0) + 4 * (-241.82 kJ/mol)] - [9.66 kJ/mol + 4 * (0)]

$\Delta$H^o = -967.28 kJ - 9.66 kJ

$\Delta$H^o = -976.94 kJ

Q4. Balanced equation : SiCl₄(l) + 2 H₂O(l) $\rightarrow$ SiO₂(s) + 4 HCl(g)

Standard enthalpy change, $\Delta$ H^o = $\Delta$ H^o_f products - $\Delta$ H^o_f reactants

$\Delta$H^o = [$\Delta$H^o_f SiO₂(s) + 4 * $\Delta$ H^o_f HCl(g)] - [$\Delta$H^o_f SiCl₄(l) + 2 * $\Delta$ H^o_f H₂O(l)]

$\Delta$H^o = [(-910.9 kJ/mol) + 4 * (-92.30 kJ/mol)] - [(-640.1 kJ/mol) + 2 * (-285.83 kJ/mol)]

$\Delta$H^o = -1280.1 kJ + 1211.76 kJ

$\Delta$H^o = -68.34 kJ