Consider these reactions, where M represents a generic metal.
2M(s)+6HCl(aq)⟶2MCl3(aq)+3H2(g)Δ𝐻1=−740.0 kJ
HCl(g)⟶HCl(aq) Δ𝐻2=−74.8 kJ
H2(g)+Cl2(g)⟶2HCl(g) Δ𝐻3=−1845.0 kJ
MCl3(s)⟶MCl3(aq) Δ𝐻4=−254.0 kJ
Use the given information to determine the enthalpy of the reaction
2M(s)+3Cl2(g)⟶2MCl3(s)
To determine the enthalpy of the reaction:
2M(s) + 3Cl2(g) ⟶ 2MCl3(s)
we need to use the given information and apply the Hess's Law, which states that the change in enthalpy for a reaction is the same regardless of the multiple steps taken to reach the final reaction.
Let's break down the desired reaction into a series of steps:
Step 1:
2M(s) + 6HCl(aq) ⟶ 2MCl3(aq) + 3H2(g)
ΔH1 = -740.0 kJ (given)
Step 2:
2MCl3(aq) ⟶ 2MCl3(s)
ΔH4 = -254.0 kJ (given)
Step 3:
HCl(g) ⟶ HCl(aq)
ΔH2 = -74.8 kJ (given)
Step 4:
H2(g) + Cl2(g) ⟶ 2HCl(g)
ΔH3 = -1845.0 kJ (given)
Now, we can manipulate these equations to cancel out the necessary species to obtain the desired reaction:
Step 1:
2M(s) + 6HCl(aq) ⟶ 2MCl3(aq) + 3H2(g)
(Multiply by 2)
(-2)M(s) - 6HCl(aq) ⟶ (-2)MCl3(aq) - 3H2(g)
Step 2:
(-2)MCl3(aq) + 3H2(g) ⟶ (-2)MCl3(s) + 3H2(g)
(Cancel out the 3H2(g) on both sides)
(-2)MCl3(aq) ⟶ (-2)MCl3(s)
Step 3:
(-2)MCl3(aq) ⟶ (-2)MCl3(aq) + 2M(s)
(Add (-2)MCl3(aq) to both sides)
0 ⟶ 2M(s)
Step 4:
HCl(aq) ⟶ HCl(aq)
(Cancel out)
Finally, we can add up the ΔH values for the steps to get the overall ΔH for the desired reaction:
ΔHdesired = ΔH1 + ΔH2 + ΔH3 + ΔH4
= -740.0 kJ + (-74.8 kJ) + (-1845.0 kJ) + (-254.0 kJ)
= -2913.8 kJ
Therefore, the enthalpy of the reaction:
2M(s) + 3Cl2(g) ⟶ 2MCl3(s)
is -2913.8 kJ.