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Asked: 2022-04-13T06:25:17+00:00 In:

Consider these reactions, where M represents a generic metal.

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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)

♦ Relevant knowledge

The enthalpy in reaction is the amount of energy in an uncontained system that does not require exchange of energy with the environment. It’s a measure of the exothermic or endothermic character of the process.

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    Answer:

    Explanation:

    Hess’ Law state:

    A chemical equation can be described as the sum or a subset of many other chemical equations. The enthalpy difference of the first chemical formula equals the sum and enthalpy differences of all the 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) DH0 = -909.0 kJ ——-> equation (1)

    HCl (g)HCl (aq) DH = 74.8 kJ ——-> (2)

    H 2 (g)+Cl 2(g]2HCl(g). DH _ 0 = +1845.0 kJ ——-> formula (3)

    MCl 3(s),MCl 3(aq), DH 0 = 398 kJ ——-> equation

    Our target equation is    2M(s)+3Cl2(g)⟶2MCl3(s)

    Step 2:

    We multiply Equation (3) by 2 Cl 2 because our target equation has 2 Cl 2.

    3x [H 2 (g) + Cl 2(g),2HCl(g),DH = -1845.0kJ

    3H2(g) + 3Cl2(g)⟶6HCl(g ΔH0 = -5535 kJ —–equation(5)

    We multiply Equation (2) with 6 to cancel equation 1.

    6x [HCl(g),HCl (aq) DH = 74.8 kJ

    6HCl(g)⟶ 6HCl(aq)   ΔH0 = -448.8 kJ ——equation (6)

    Our target equation does not have MCl 3(aq). Therefore, we must reverse equation 4 in order to cancel the MCl 3(aq), and multiply it by 2.

    2 x [MCl (3(aq),MCl (3(s), DH = +398 KJ ]

    [note: sign will reversed so it becomes +ve]

    2 MCl3(aq)⟶ 2 MCl3(s)   ΔH0 = +796  kJ —-equation (7)

    Step 3:

    Add equation (1). To get the target equation, add equations (1), (6), and (7)

    2M(s)+6HCl(aq)2MCl3(aq)+3H2(g) DH0 = -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)

    Add all of the above to 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)

    DH 0 = 909.0 kJ + (+5535 kJ) + (+448.8.8 kJ) + +796.kJ = -6096.8 kJ

    Similar species of the same kind will cancel each other (here  6HCl(g),  3H2(g) , 2MCl3(aq) , 6HCl(aq) will cancel ) from both sides of reaction

    The overall equation will become

    2M(s)+3Cl2(g)⟶2MCl3(s) DH0 = -6096.8 kJ

    Hence, enthalpy of the reaction  is -6096.8 kJ   

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