Consider these reactions, where M represents a generic metal.

Question

2M(s)+6HCl(aq)⟶2MCl3(aq)+3H2(g)Δ𝐻1=−901.0 kJ
HCl(g)⟶HCl(aq) Δ𝐻2=−74.8 kJ
H2(g)+Cl2(g)⟶2HCl(g) Δ𝐻3=−1845.0 kJ
MCl3(s)⟶MCl3(aq) Δ𝐻4=−195.0 kJ
Use the given information to determine the enthalpy of the reaction

2M(s)+3Cl2(g)⟶2MCl3(s)

Answer 1

To determine the enthalpy of the reaction 2M(s) + 3Cl2(g) ⟶ 2MCl3(s), we can use Hess's Law.

Hess's Law states that the enthalpy change of a reaction is independent of the pathway taken and depends only on the initial and final states of the reaction. Therefore, we can manipulate the given reactions to obtain the desired reaction and then sum up their enthalpy changes.

First, we need to cancel out any common compounds that appear on both sides of the equations. We can cancel out HCl(aq) from reaction 1 and HCl(aq) from reaction 4, leaving us with:

2M(s) + 6HCl(aq) ⟶ 2MCl3(aq) + 3H2(g) (equation 1)
HCl(g) ⟶ HCl(aq) (equation 2)
H2(g) + Cl2(g) ⟶ 2HCl(g) (equation 3)
MCl3(s) ⟶ MCl3(aq) (equation 4)

Next, we can manipulate these equations to obtain the desired reaction:

Multiply equation 1 by -1:
-2M(s) - 6HCl(aq) ⟶ -2MCl3(aq) - 3H2(g)

Multiply equation 2 by -3:
-3HCl(g) ⟶ -3HCl(aq)

Multiply equation 3 by 2:
2H2(g) + 2Cl2(g) ⟶ 4HCl(g)

Finally, we can sum up these equations to obtain our desired reaction:

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

Now, we can sum up the enthalpy changes of these equations to determine the enthalpy of the desired reaction:

ΔH = ΔH1 + ΔH2 + ΔH3 + ΔH4

ΔH = -901.0 kJ + (-74.8 kJ) + (-1845.0 kJ) + (-195.0 kJ)

ΔH = -901.0 kJ - 74.8 kJ - 1845.0 kJ - 195.0 kJ

ΔH = -3015.8 kJ

Therefore, the enthalpy of the reaction 2M(s) + 3Cl2(g) ⟶ 2MCl3(s) is -3015.8 kJ.