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It-Gec-H(g) - 2HK(g)-1-(-f-r-a) Bond: CIC C-C H-I C-I C-H /mol): 839 348 299 240 413 Using the table of average bond energies below the image, tr Delta H for the reaction in kJis __ +160kJ -160kJ -63kJ -217kJ

Problemas

It-Gec-H(g) - 2HK(g)-1-(-f-r-a)
Bond: CIC C-C H-I C-I C-H
/mol): 839 348 299 240 413
Using the table of average bond energies below the image, tr
Delta H for the reaction in kJis __
+160kJ
-160kJ
-63kJ
-217kJ

It-Gec-H(g) - 2HK(g)-1-(-f-r-a) Bond: CIC C-C H-I C-I C-H /mol): 839 348 299 240 413 Using the table of average bond energies below the image, tr Delta H for the reaction in kJis __ +160kJ -160kJ -63kJ -217kJ

Solución

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Soniaveterano · Tutor durante 10 años
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To calculate the enthalpy change (ΔH) for the reaction, we need to consider the bond energies of the bonds broken and formed during the reaction.<br /><br />Given information:<br />- Bond energies (in kJ/mol):<br /> - C-Cl: 339 kJ/mol<br /> - C-C: 839 kJ/mol<br /> - H-I: 299 kJ/mol<br /> -I: 413 kJ/mol<br /> - C-H: 348 kJ/mol<br /><br />The reaction is:<br />C2Cl2 + 2HI → 2C2H + 2I2<br /><br />Bonds broken:<br />- 2 C-Cl bonds: 2 × 339 kJ/mol = 678 kJ/mol<br />- 2 H-I bonds: 2 × 299 kJ/mol = 598 kJ/mol<br />Total energy of bonds broken = 678 kJ/mol + 598 kJ/mol = 1,276 kJ/mol<br /><br />Bonds formed:<br />- 4 C-H bonds: 4 × 348 kJ/mol = 1,392 kJ/mol<br />- 2 C-I bonds: 2 × 413 kJ/mol = 826 kJ/mol<br />Total energy of bonds formed = 1,392 kJ/mol + 826 kJ/mol = 2,218 kJ/mol<br /><br />Enthalpy change (ΔH) = Energy of bonds formed - Energy of bonds broken<br />ΔH = 2,218 kJ/mol - 1,276 kJ/mol = 942 kJ/mol<br /><br />Therefore, the correct answer is:<br />$+160kJ$
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