Student Textbook Questions page 681 - Hess’s Law
11. Ethene, C2H4, reacts with water to form ethanol, CH3CH2OH(l) .
C2H4 (g) + H2O(l) → CH3CH2OH(l)
Determine the enthalpy change of this reaction, given the following thermochemical equations.
(1) CH3CH2OH(l) + 3O2 (g) → 3 H2O(l) + 2CO2(g) ∆H ° =-1367 kJ/mol
(2) C2H4 (g) + 3O2 (g) → 2 H2O(l)) + 2CO2(g) ∆H ° = -1411 kJ/mol
Plan Your Strategy
Step 1: Examine equations (1) and (2) to see how they compare with the target equation. Decide how you need
to manipulate equations (1) and (2) so that they add to give the target equation. (Reverse the equation, multiply
the equation, do both, or do neither.) Remember to adjust ∆H ° accordingly for each equation.
Step 2: Write the manipulated equations so that their equation arrows line up. Add the reactants and products
on each side, and cancel substances that appear on both sides.
Step 3: Ensure that you have obtained the target equation. Add ∆H ° for the combined equations.
Solution:
i) Equation (1) has 1 mol CH3CH2OH(l) as a reactant. This is required as a product, so equation (1) needs to be
“flipped,” or multiplied by –1:
3 H2O(l) + 2CO2(g) → CH3CH2OH(l) + 3O2 (g)
∆H ° = + 1367 kJ/mol
ii) Equation (2) has 1 mol C2H4 (g) as a reactant, as required, so it is left as is.
C2H4 (g) + 3O2 (g) → 2 H2O(l)) + 2CO2(g) ∆H ° = -1411 kJ/mol
iii) Add the resulting equations
3 H2O(l) + 2CO2(g) + C2H4 (g) + 3O2 (g) → CH3CH2OH(l) + 3O2 (g) + 2 H2O(l) + 2CO2(g)
H2O(l) + C2H4 (g) → CH3CH2OH(l) which matches the target equation.
iv) Add the resulting ∆H °
∆H ° = ∆H °1 + ∆H °2 = +1367 kJ +(-1411 kJ) = -44 kJ
12. A typical automobile engine uses a lead acid battery. During discharge, the following chemical reaction
takes place: Pb(s) +PbO2(s) +2H2SO4(l) → 2PbSO4(s) +2H2O(l)
Determine the enthalpy change of this reaction, given the following equations.
(1) Pb(s) +PbO2(s) +2SO3(g) → 2PbSO4(s)
(2) SO3(g) + H2O(l) → H2SO4(l)
∆H ° = -775 kJ
∆H ° = -133 kJ
Solution:
i) Equation 1 has Pb and PbO2 as reactants, and 2PbSO4 as a product, matching the target equation. Therefore,
equation 1 needs to stay the same.
Pb(s) + PbO2(s) +2SO3(g) → 2PbSO4(s)
∆H ° = -775 kJ
ii) Equation 2 has 1 mol of H2SO4 as a product. The target equation requires 2 moles of H2SO4 as a reactant.
Hence, you must multiply equation 2 by –2 (or “flip” and double).
2H2SO4(l) → 2SO3(g) + 2H2O(l)
∆H ° = 266 kJ
iii) Pb(s) + PbO2(s) + 2SO3(g) + 2H2SO4(l) → 2PbSO4(s) + 2SO3(g) + 2H2O(l)
Pb(s) + PbO2(s) + 2H2SO4(l) → 2PbSO4(s) + 2H2O(l) (same as target equation)
iv) ∆H ° = ∆H °1 + ∆H °2 = -775 kJ + 266 kJ = -509 kJ
13. Mixing household cleansers can result in the production of hydrogen chloride gas, HCl(g). Not only is this
gas dangerous in its own right, it also reacts with oxygen to form chlorine gas and water vapour: 4HCl(g) + O2(g)
→ 2 Cl2(g) + 2H2O(g) Determine the enthalpy change of this reaction, given the following equations.
(1) H2 (g) +Cl2 (g) → 2 HCl (g)
(2) H2 (g) + ½ O2 (g) → H2O (l)
(3) H2O (g) → H2O (l)
∆H ° = -185 kJ
∆H ° = -285.8 kJ
∆H ° = -40.7 kJ
Solution:
i) Equation (1) has 2HCl as a product and Cl2 as a reactant. The target equation requires 4HCl on the reactant
side and 2Cl2 on the product side (opposite). Hence, multiply equation (1) by –2.(or double and “flip”).
4HCl (g) → 2H2 (g) + 2Cl2 (g)
∆H ° = 370 kJ
ii) Equation (2) must be multiplied by 2 to obtain O2 on the reactant side.
2H2 (g) + O2 (g) → 2H2O (l)
∆H ° = 571.6 kJ
iii) Multiply equation (3) by –2 (or flip and double) to allow cancelling of 2H2O (l)
2H2O (l) → 2H2O (g)
∆H ° = 81.4 kJ
iv) 4HCl (g) + 2H2 (g) + O2 (g) ) + 2H2O (l) → 2H2 (g) + 2Cl2 (g) + 2H2O (l) + 2H2O (g)
4HCl(g) + O2(g) → 2 Cl2(g) + 2H2O(g)
v) ∆H ° = 370 kJ + 571.6 kJ + 81.4 kJ = -120 kJ
14. Calculate the enthalpy change of the following reaction between nitrogen gas and
oxygen gas, given thermochemical equations (1), (2) and (3):
2N2 (g) +5O2 (g) → 2N2O5 (g)
(1) 2H2 (g) + O2 (g) → 2H2O (l)
(2) N2O5 (g) + H2O (l) → 2HNO3 (l)
(3) ½ N2 (g) + 3/2 O2 (g) + ½ H2 (g) → HNO3 (l)
∆H ° = -572 kJ
∆H ° = -77 kJ
∆H ° = -174 kJ
Solution:
i) Equation (2) has N2O5 as a reactant. The target equation requires 2 N2O5 on the product side. Hence, multiply
equation (2) by –2. (or double and flip)
4HNO3 (l) → 2N2O5 (g) + 2H2O (l)
∆H ° = 154 kJ
ii) Equation (3) must be multiplied by 4 to give 2 mol N2 on the reactant side.
2 N2 (g) + 6 O2 (g) + 2 H2 (g) → 4 HNO3 (l)
∆H ° = - 696 kJ
iii) To allow cancelling of H2O, H2, and O2, flip equation (1) (or multiply by –1).
2H2O (l) → 2 H2 (g) + O2 (g)
∆H ° = 572 kJ
iv) 4HNO3 (l) + 2 N2 (g) + 6 O2 (g) + 2H2 (g) + 2H2O (l) → 2N2O5 (g) + 2H2O (l) + 4HNO3 (l) + 2H2 (g) + O2 (g)
2N2 (g) +5O2 (g) → 2N2O5 (g)
v) ∆H ° = 154 kJ + - 696 kJ + 572 kJ = 30 kJ