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reaction enthalpy for the formation of aluminum chloride?
can someone please help me and explain how to get the answer? the first correct answer gets a best answer. thanks a lot!
Calculate the reaction enthalpy for the formation
2 Al(s) + 3 Cl2(g) − → 2 AlCl3(s) ,
of anhydrous aluminum chloride using the data
2 Al(s) + 6 HCl(aq) − → 2 AlCl3(aq) + 3 H2(g) ∆H◦ = −1049 kJ
HCl(g) −→ HCl(aq) ∆H◦ = −74.8 kJ
H2 (g) + Cl2 (g) −→ 2 HCl(g) ∆H◦ = −185 kJ
AlCl3(s) −→ AlCl3(aq) ∆H◦ = −323 kJ
1. −1450.85 kJ
2. −1826.2 kJ
3. −1502.4 kJ
4. −1883.5 kJ
5. −1225.7 kJ
6. −1100.36 kJ
7. −1406.8 kJ
3 Answers
- 1 decade agoFavorite Answer
2 Al(s) + 6 HCl(aq) >> 2 AlCl3(aq) + 3 H2(g)…|……-1049
+ 6 x HCl(g) >> HCl(aq)…………………………|…+ 6x (-74.8)
+ 3 x H2(g) + Cl2(g) >> 2 HCl(g).………………|…+ 3x (-185)
- 2 x AlCl3(s) >> AlCl3(aq)………………………|…- 2x (-323)
————————————————————————————
2 Al(s) + 6 HCl(aq) + 6 HCl(g) + 3 H2(g) + 3 Cl2(g) - 2 AlCl3(s)
…>>…
2 AlCl3(aq) + 3 H2(g) + 6 HCl(aq) + 6 HCl(g) - 2 AlCl3(aq)
…
reducing (?) delivers:
2 Al(s) + 3 Cl2(g) >> 2 AlCl3(s)
.……………………= -1049 - 448.8 - 555 + 646 = -1406.8 kJ
Source(s): Just what I learned during studies - AprilLv 45 years ago
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