1. 2M(s)+6HCl(aq) \Rightarrow 2MCl3(aq)+3H2(g)2M(s)+6HCl(aq) \Rightarrow 2MCl3(aq)+3H2(g) \triangle H_{1}=-819.0kJ 2. HCl(g) \Rightarrow HCl(aq)HCl(g)

Jean Blumer 2022-01-10 Answered
1. 2M(s)+6HCl(aq)2MCl3(aq)+3H2(g)2M(s)+6HCl(aq)2MCl3(aq)+3H2(g)
H1=819.0kJ
2. HCl(g)HCl(aq)HCl(g)HCl(aq)
H2=74.8kJ
3. H2(g)+Cl2(g)2HCl(g)H2(g)+Cl2(g)2HCl(g)
H3=1845.0kJ
4. MCl3(s)MCl3(aq)
H4=128.0kJ
Use the given information to determine the enthalpy of the reaction
2M(s)+3Cl2(g)2MCl3(s)
H=?
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Expert Answer

Melinda McCombs
Answered 2022-01-11 Author has 38 answers
Step 1
In the above reactions we only need 2 reactions to calculate the enthalpy of the following reaction;
2M(s)+3Cl2(g)MCl3(g)
Step 2
We have 4 chemical equations from which 2 are important:
2M(s)+6HCl(aq)2MCl3(aq)+3H2(g)
H1=819KJ
H2(g)+Cl2(g)2HCl(g)
H3=1845KJ
Multiplying equation 1st by 3
3H2(g)+3Cl2(g)6HCl(g)
H3=3×(1845KJ)=5535KJ
Adding equation 1st and 3rd
2M(s)+3Cl2(g)2MCl3(aq)
Hreaction
=(H1+H3)=4716KJ
Therefore the enthalpy of formation =4716KJ (answer)
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Ben Owens
Answered 2022-01-12 Author has 27 answers
Chemical reaction 1: 2M(s)+6HCl(aq)2MCl3(aq)+3H2(g);H1=556.0kJ. Chemical reaction 2: HCl(g)HCl(aq);H2=74.8kJ.
Chemical reaction 3: H2(g)+Cl2(g)2HCl(g);H3=1845.0kJ
Chemical reaction 4: MCl3(s)MCl3(aq);H4=342.0kJ.
Chemical reaction 5: 2M(s)+3Cl2(g)2MCl3(s);H5=?.H5=H1+6·H2+3H32·H4.
H5=550kJ+6·(74,8kJ)+3(1845kJ)2(342kJ).
H5=550kJ448,8kJ5535kJ+684kJ.
H5=5849,8kJ.
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star233
Answered 2022-01-15 Author has 137 answers

On adding,
2M(s)+3Cl2(g)2MCl3(s)H=944kJ+6×(74.8kJ)+3×(1845kJ)+(2)×(234kJ)H=944kJ448.85535kJ+468kJH=6459.8kJ

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