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Consider these reactions, where M represents a generic metal.

2M(s)+6HCl(aq)⟶2MCl_{3}(aq)+3H_{2}(g)Δ?1=−909.0 kJ

HCl(g)⟶HCl(aq) Δ?2=−74.8 kJ

H_{2}(g)+Cl_{2}(g)⟶2HCl(g) Δ?3=−1845.0 kJ

MCl_{3}(s)⟶MCl_{3}(aq) Δ?4=−398.0 kJ

Use the given information to determine the enthalpy of the reaction 2M(s)+3Cl_{2}(g)⟶2MCl_{3}(s)

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.

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 reversesif a chemical reaction is multiply by any number then the enthalpy of the reaction is also multiply by that same numberStep 1: write all the reactions2M(s)+6HCl(aq)2MCl

_{3}(aq)+3H_{2}(g) DH^{0}= -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 ——-> equationOur target equation is 2M(s)+3Cl_{2}(g)⟶2MCl_{3}(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 = -1845.0kJ3H_{2}(g) + 3Cl_{2}(g)⟶6HCl(g ΔH^{0}= -5535 kJ —–equation(5)We multiply Equation (2) with 6 to cancel equation 1.

6x [HCl(g),HCl (aq) DH

^{}= 74.8 kJ6HCl(g)⟶ 6HCl(aq) ΔH^{0}= -448.8 kJ ——equation (6)Our target equation does not have MCl

_{3}(aq). Therefore, we must reverse equation4in 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 MCl_{3}(aq)⟶ 2 MCl_{3}(s) ΔH^{0}= +796 kJ —-equation (7)Step 3:Add equation (1). To get the target equation, add equations (1), (6), and (7)

2M(s)+6HCl(aq)2MCl

_{3}(aq)+3H_{2}(g) DH^{0}= -909.0 kJ —-equation (1)3H_{2}(g) + 3Cl_{2}(g)⟶6HCl(g) ΔH^{0}= -5535 kJ —–equation(5)6HCl(g)⟶ 6HCl(aq) ΔH^{0}= -448.8 kJ ——equation (6)2 MCl_{3}(aq)⟶ 2 MCl_{3}(s) ΔH^{0}= +796 kJ —-equation (7)Add all of the above to get

2M(s)+6HCl(aq) + 3H

_{2}(g) + 3Cl_{2}(g) + 6HCl(g) + 2 MCl_{3}(aq) ———–> 2MCl_{3}(aq)+3H_{2}(g) + 6HCl(g) + 6HCl(aq) + 2 MCl_{3}(s)DH

^{0}= 909.0 kJ + (+5535 kJ) + (+448.8.8 kJ) + +796.kJ =-6096.8 kJSimilar species of the same kind will cancel each other (here

6HCl(g), 3Hwill cancel ) from both sides of reaction_{2}(g) , 2MCl_{3}(aq) , 6HCl(aq)The overall equation will become

2M(s)+3ClDH_{2}(g)⟶2MCl_{3}(s)^{0}=-6096.8 kJHence, enthalpy of the reaction is -6096.8 kJ