Question

Consider these reactions, where M represents a generic metal.

Answer 1

Consider these reactions, where M represents a generic metal.  Determine ΔH₅.

1. 2M (s) + 6HCl (aq) ⟶ 2MCl₃ (aq) + 3H₂ (g); ΔH₁ = −878.0 kJ
2. HCl (g) ⟶ HCl (aq); ΔH₂ = −74.8 kJ
3. H₂ (g) + Cl₂ (g) ⟶ 2HCl (g); ΔH₃ = −1845.0 kJ
4. MCl₃ (s) ⟶ MCl₃ (aq); ΔH₄ = −497.0 kJ
5. 2M (s) + 3Cl₂ (g) ⟶ 2MCl₃ (s); ΔH₅ = ? kJ

We need to manipulate the equations on top to output the equation on the bottom.  Hess's Law states that the total enthalpy change of the desired reaction is the sum of all ethalpy changes:

Δ$H_{overall}=H_{1} +H_{2} +H_{3} +...+H_{n}$.

We will need to arrange the equations so that the compounds/ elements that are not in the desired reaction will cancel out.  This can be done by reversing the reactions, and by multiplying or dividing a given equation by a factor.

For reactants in our desired reaction we have 2M (s) and 3Cl₂ (g) and their product is 2MCl₃ (s), so the compounds/ elements in the equations that will need to cancel are HCl (aq), H₂ (g), HCl (g), and MCl₃ (aq).

• HCl (aq).  The HCl (aq)'s are on opposite sides of the arrow, but they have different coefficients.  To be able to cancel, you will need to multiply the second HCl (aq) by 6.  Just like in algebra, you have to multiply the entire equation, not just one variable.  Also, since you're multiplying the equation by 6, its enthalpy will also increase by a factor of 6.

2. 6(HCl (g) ⟶ HCl (aq)); ΔH₂ = 6(−74.8) kJ

2. 6HCl(g) ⟶ 6HCl (aq); ΔH₂ = -448.8 kJ

• H₂ (g).  The equations with hydrogen gas are the first and the third.  The first hydrogen has a coefficient of 3, and the second, a coefficient of 1.  Multiply the second equation and its enthalpy by 3:

3. 3(H₂ (g) + Cl₂ (g) ⟶ 2HCl (g)); ΔH₃ = 3(−1845.0) kJ

3. 3H₂ (g) + 3Cl₂ (g) ⟶ 6HCl (g); ΔH₃ = -5535.0 kJ

• HCl (g).  The equations containing HCl in a gaseous form are equations 2 and 3, which we already manipulated for other variables, but in the process we made kept the HCl (g)'s the same.

2. 6HCl(g) ⟶ 6HCl (aq); ΔH₂ = -448.8 kJ

3. 3H₂ (g) + 3Cl₂ (g) ⟶ 6HCl (g); ΔH₃ = -5535.0 kJ

• MCl₃ (aq).  In equation 1, MCl₃ (aq) has a coefficient of 2.  Multiply equation 4 and its enthalpy by 2.  To cancel terms, you will need to reverse the equation.  This changes the sign on the enthalpy from negative to positive.

4. 2(MCl₃ (s) ⟶ MCl₃ (aq)); ΔH₄ = 2(−497.0) kJ

4. 2MCl₃ (s) ⟶ 2MCl₃ (aq); ΔH₄ = -994.0 kJ

4. 2MCl₃ (aq) ⟶ 2MCl₃ (s); ΔH₄ = 994.0 kJ

Now when you add all the products and the reactants, the resulting equation should be 2M (s) + 3Cl₂ (g) ⟶ 2MCl₃ (s) (underlined terms cancel since they are on both sides):

2M (s) + 6HCl (aq) ⟶ 2MCl₃ (aq) + 3H₂ (g)

6HCl(g) ⟶ 6HCl (aq)

3H₂ (g) + 3Cl₂ (g) ⟶ 6HCl (g)

2MCl₃ (aq) ⟶ 2MCl₃ (s)

2M (s) + 6HCl (aq) + 6HCl(g) + 3H₂ (g) + 3Cl₂ (g) + 2MCl₃ (aq) ⟶ 2MCl₃ (aq) + 3H₂ (g) + 6HCl (aq) + 6HCl (g) + 2MCl₃ (s)

2M (s) + 3Cl₂ (g) ⟶ 2MCl₃ (s)

Finally, add the enthalpies:

−878.0 kJ + (-448.8 kJ) + (-5535.0 kJ) + 994.0 kJ = -5869 kJ

Answer:

ΔH₅ = -5869 kJ