Ejercicios Ley De Hess

Actividad 1.3

1. N2(g) + ½ O2(g) | | N2O(g) | | ∆H°ʀ=? |
a) 3O2(g) + 4NH3(g) | | 6H2O(I) + 2N2(g) | (1/6) | ∆H°ʀ= -365.9 Kcal |
b) 3N2O(g) + 2NH3(g) | | 4N2(g) + 3H2O(I) | (1/3) | ∆H°ʀ= -241.39 Kcal |
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c) 4/3N2 + H2O(I) | | N2O(g) + 2/3 NH3(g) | | ∆H°ʀ= 80.46 Kcal |
d) ½ O2 + 2/3NH3(g) | | H2O(I) + 1/3N2(g) | | ∆H°ʀ= -60.98 Kcal |
N2(g) + ½O2(g) | | N2O(g) | | ∆H°ʀ= 19.48 Kcal |

2. Cu(S) + 2NF3(g) | | CuF2(s) + N2F4(g) | | ∆H°ʀ=? |
a) Cu(S) + F2(g) | | CuF2(s) | | ∆H°ʀ=-126.91Kcal |
b)2NO(g) + 2NF3(g) | | 2ONF(g) + N2F4(g) | | ∆H°ʀ=-19.81Kcal |
c)NO(g) + 1/2F(2) | | ONF(g) | (-2) | ∆H°ʀ= -32.5Kcal |
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d) 2NO(g) + 2NF3(g) | | 2ONF(g) + N2F4(g) | | ∆H°ʀ=-19.81Kcal |
e) Cu(S) + F2(g) | |CuF2(s) | | ∆H°ʀ=-126.91Kcal |
f) 2ONF(g) | | 2NO(g) + F(2) | | ∆H°ʀ=65Kcal |
Cu(S) + 2NF3(g) | | CuF2(s) + N2F4(g) | | ∆H°ʀ=-81.72Kcal |
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3. 4CIF3(g) + 3N2H4(I) | | 2Cl2(g) + 12HF(g) + 3N2(g) | | ∆H°ʀ=? |
a) 4NH3(g) + 3O2(g) | | 6H2O(I) + 2N2(g) | (-1) | ∆H°ʀ=-365.82Kcal |
b)2NH3(g) + 2CIF3(g) | | Cl2(g) + N2(g) + 6HF(g) | (2) | ∆H°ʀ=-285.75Kcal |
c)O2(g) +N2H4(I) | | N2(g) + 2H2O(I) | (3) | ∆H°ʀ=-148.75Kcal |
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d) 4NH3(g) + 4CIF3(g) | | 2Cl2(g) + 2N2(g) + 12HF(g) | | ∆H°ʀ=-571.5Kcal |
e) 3O2(g) + 3N2H4(I) | | 3N2(g) + 6H2O(I) | | ∆H°ʀ=-446.25Kcal |
f) 6H2O(I) + 2N2(g) | | 4NH3(g) + 3O2(g) | | ∆H°ʀ=365.82Kcal |
4CIF3(g) + 3N2H4(I) | | 2Cl2(g) + 12HF(g) + 3N2(g) | | ∆H°ʀ=-651.93Kcal |

4. Na+1(g) + Cl-1(g) | | NaCl(S) || ∆H°ʀ=? |
a) Cl2(g) | | 2Cl(g) | (-1/2) | ∆H°ʀ=58.03Kcal |
b)Na(S) + 1/2Cl2(g) | | NaCl(S) | | ∆H°ʀ=-98.255Kcal |
c)Na(S) | | Na(g) | (-1) | ∆H°ʀ=25.9799Kcal |
d) Cl-1(g) | | Cl(g) + 1e- | | ∆H°ʀ=87.308Kcal |
e) Na(g) | | Na+1(g) + 1e- | (-1) | ∆H°ʀ=120Kcal |
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f) Na(S) + 1/2Cl2(g) | | NaCl(S) | | ∆H°ʀ=-98.255Kcal |
g) Cl-1(g) | | Cl(g) + 1e- | |∆H°ʀ=87.308Kcal |
h) Na+1(g) + 1e- | | Na(g) | | ∆H°ʀ=-120Kcal |
i) Na(g) | | Na(S) | | ∆H°ʀ=-25.979Kcal |
j) Cl(g) | | 1/2Cl2(g) | | ∆H°ʀ=-29.015Kcal |
Na+1(g) + Cl-1(g) | | NaCl(S) | | ∆H°ʀ=-185.94Kcal |

5. S(S) + 2 O2(g) + H2(g) | | H2SO4(I) | | ∆H°ʀ=? |
a) S(S) + O2(g) | | SO2(g) | | ∆H°ʀ=-71.0Kcal |
b)2H2O(I) | | O2 + 2H2(g) | (-1/2) | ∆H°ʀ=136.64Kcal |
c) SO3](g) + H2O(I) || H2SO4(I) | | ∆H°ʀ=-31.14Kcal |
d)2SO3(g) | | 2SO(g) + O2(g) | (-1/2) | ∆H°ʀ=46.98Kcal |
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e) SO3](g) + H2O(I) | | H2SO4(I) | | ∆H°ʀ=-31.14Kcal |
f) 1/2O2 + H2(g) | | H2O(I) | | ∆H°ʀ=-68.32Kcal |
g) SO(g) + 1/2O2(g) | | SO3(g) | | ∆H°ʀ=-23.49Kcal |
h) S(S) + O2(g) | | SO2(g) | | ∆H°ʀ=-71.0Kcal |
S(S) + 2 O2(g) + H2(g) | | H2SO4(S) | | ∆H°ʀ=-193.95Kcal |6.P(S)+1/2O2(g) +3/2Cl2(g) | | POCl3(I) | | ∆H°ʀ=? |
a) P(S) + 5/2Cl2(g) | | PCl5(I) | | ∆H°ʀ=-95.4Kcal |
b)PCl5(I) + H2O(I) | | 2HCl(I) + POCl3(I) | | ∆H°ʀ=-32.5Kcal |
c) 1/2H2(g) + 1/2Cl2(g) | | HCl(I) | (-2) | ∆H°ʀ=-22.1Kcal |
d)H2(g) + ½ O2(g) | | H2O(I) | | ∆H°ʀ=-68.32Kcal |
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e) PCl5(I) + H2O(I) | | 2HCl(I) + POCl3(I) | | ∆H°ʀ=-32.5Kcal |
f) P(S) +5/2Cl2(g) | | PCl5(I) | | ∆H°ʀ=-95.4Kcal |
g) H2(g) + ½ O2(g) | | H2O(I) | | ∆H°ʀ=-22.1Kcal |
h) 2HCl(I) | | H2(g) + Cl2(g) | | ∆H°ʀ=-68.32Kcal |
P(S)+1/2O2(g) +3/2Cl2(g) | | POCl3(I) | | ∆H°ʀ=-218.32Kcal |

7. 2H3BO3(aq) | | B2O3(S) + 3H2O(I) | | ∆H°ʀ=? |
a) H2B4O7(I) | | 2B2O3(S) + H2O(I) | (1/2) | ∆H°ʀ=73.22Kcal |
b)HBO2(I) + H2O(I) | | H3BO3(aq) | (-2) | ∆H°ʀ=0.0837Kcal |C) H2B4O7(I) + H2O(g) | | 4HBO2(I) | (-1/2) | ∆H°ʀ=-47.279Kcal |
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d) 2H3BO3(aq) | | 2HBO2(I) + 2H2O(I) | | ∆H°ʀ=0.1674Kcal |
e) 1/2H2B4O7(I) | | B2O3(S) + 1/2H2O(I) | | ∆H°ʀ=36.61Kcal |
f) 2HBO2(I) | | 1/2H2B4O7(I) + 1/2H2O(I) | | ∆H°ʀ=23.6395Kcal |
2H3BO3(aq) | | B2O3(S) + 3H2O(I) | | ∆H°ʀ=60.4169Kcal |

8. Calcular la formación del C4H10 (g) a partir...