Question

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 1

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:  

• if a chemical reaction is reversed then the sign of the enthalpy of the reaction reverses
• if a chemical reaction is multiply by any number  then the enthalpy of the reaction is also multiply by that same number

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