Consider these reactions, where M represents a generic metal.
2M(s)+6HCl(aq)⟶2MCl3(aq)+3H2(g)Δ?1=−909.0 kJ
HCl(g)⟶HCl(aq) Δ?2=−74.8 kJ
H2(g)+Cl2(g)⟶2HCl(g) Δ?3=−1845.0 kJ
MCl3(s)⟶MCl3(aq) Δ?4=−398.0 kJ
Use the given information to determine the enthalpy of the reaction 2M(s)+3Cl2(g)⟶2MCl3(s)
Answer:
Explanation:
Hess' Law state:
If a chemical equation can be written as the sum of several other chemical equations, the enthalpy change of the first chemical equation equals the sum of the enthalpy changes of the other chemical equations.
Note:
Step 1: write all the reactions
2M(s)+6HCl(aq)⟶2MCl3(aq)+3H2(g) ΔH0 = -909.0 kJ -------> equation (1)
HCl(g)⟶HCl(aq) ΔH0 = -74.8 kJ -------> equation (2)
H2(g)+Cl2(g)⟶2HCl(g) ΔH0 = -1845.0 kJ -------> equation (3)
MCl3(s)⟶MCl3(aq) ΔH0 = -398 kJ -------> equation (4)
Our target equation is 2M(s)+3Cl2(g)⟶2MCl3(s)
Step 2:
Our target equation has 2 Cl2 so we multiply Equation (3) by 3
3× [ H2(g) + Cl2(g)⟶2HCl(g) ΔH0 = -1845.0 kJ ]
3H2(g) + 3Cl2(g)⟶6HCl(g ΔH0 = -5535 kJ -----equation(5)
Our target equation has no HCl (aq) so we multiply Equation (2) by 6 so it will cancel out with equation 1
6× [ HCl(g)⟶HCl(aq) ΔH0 = -74.8 kJ ]
6HCl(g)⟶ 6HCl(aq) ΔH0 = -448.8 kJ ------equation (6)
Now our target equation has no MCl3(aq) , so we need toreverse equation 4 to cancel the MCl3(aq) and multiply it by 2
2 × [ MCl3(aq)⟶MCl3(s) ΔH0 = +398 kJ ] [note: sign will reversed so it becomes +ve ]
2 MCl3(aq)⟶ 2 MCl3(s) ΔH0 = +796 kJ ----equation (7)
Step 3:
Now add equation (1) .(5), (6) and (7) to get the target equation
2M(s)+6HCl(aq)⟶2MCl3(aq)+3H2(g) ΔH0 = -909.0 kJ ----equation (1)
3H2(g) + 3Cl2(g)⟶6HCl(g) ΔH0 = -5535 kJ -----equation(5)
6HCl(g)⟶ 6HCl(aq) ΔH0 = -448.8 kJ ------equation (6)
2 MCl3(aq)⟶ 2 MCl3(s) ΔH0 = +796 kJ ----equation (7)
On adding the all the above equation we get
2M(s)+6HCl(aq) + 3H2(g) + 3Cl2(g) + 6HCl(g) + 2 MCl3(aq) -----------> 2MCl3(aq)+3H2(g) + 6HCl(g) + 6HCl(aq) +
2 MCl3(s)
ΔH0 = -909.0 kJ + ( -5535 kJ ) + (-448.8 kJ ) + +796 kJ = -6096.8 kJ
Now, similar like species will cancel out ( here 6HCl(g), 3H2(g) , 2MCl3(aq) , 6HCl(aq) will cancel out ) from each side of reaction
So overall equation will becomes
2M(s)+3Cl2(g)⟶2MCl3(s) ΔH0 = -6096.8 kJ
Hence, enthalpy of the reaction is -6096.8 kJ
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