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zloy xaker [14]
3 years ago
12

7. A nonpolar covalent bond (i.e., pure covalent) would form in which one of the following pairs of atoms? A) Na Cl B) H Cl C) L

i Br D) Se Br E) Br Br
Chemistry
1 answer:
padilas [110]3 years ago
8 0

Answer:

E) Br Br

Explanation:

Covalent bond -

Covalent bond is the bond which is formed with the sharing of the electrons between the two atoms which are taking part in the bond. It is generally formed between the atoms with similar electronegativity values.

It is the bond which is generally occurs within non metals as they share electrons to complete their octet.

The difference in the electronegativity values of the atoms involving in a covalent bond must not exceed the value of 1.7 .

A covalent bond to be non polar in nature , then the atoms need to have almost similar electronegativity , so that they won't split into partial negative and partial positive charge .

Hence , from the given options of the question ,

E) Br Br , both the atoms are same , hence same electronegativity , therefore are non - polar in nature .

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Standard Heat of Reaction 1 = -136.2 kJ/mol

Standard Heat of Reaction 2 = -41.166 kJ/mol

Standard Heat of Reaction 3 = -136.07 kJ/mol

Standard Heat of Reaction 4 = 279.448kJ/mol

Explanation:

C₂H₄ (g) + H₂ (g) → C₂H₆ (g)

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H₂ (g), 0 kJ/mol

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H₂O (g), -241.818 kJ/mol

H₂ (g), 0 kJ/mol

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NO₂ (g), 33.2 kJ/mol

H₂O (l), -285.8 kJ/mol

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NO (g), 90.29 kJ/mol

Cr₂O₃ (s), -1128.4 kJ/mol

CO (g), -110.525 kJ/mol

Cr (s), 0 kJ/mol

CO₂ (g), -393.509 kJ/mol

Note that

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

C₂H₄ (g) + H₂ (g) → C₂H₆ (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (1×-83.7) = -83.7 kJ/mol

ΔH∘(reactants) = (1×52.5) + (1×0) = 52.5 kJ/mol

ΔH∘(rxn) = -83.7 - 52.5 = -136.2 kJ/mol

CO (g) + H₂O (g) → H₂ (g) + CO₂ (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (1×0) + (1×-393.509) = -393.509 kJ/mol

ΔH∘(reactants) = (1×-110.525) + (1×-241.818) = -352.343 kJ/mol

ΔH∘(rxn) = -393.509 - (-352.343) = -41.166 kJ/mol

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ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

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ΔH∘(rxn) = -322.27 - (-186.2) = -136.07 kJ/mol

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ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (2×0) + (3×-393.509) = -1,180.527 kJ/mol

ΔH∘(reactants) = (1×-1128.4) + (3×-110.525) = -1,459.975 kJ/mol

ΔH∘(rxn) = -1,180.527 - (-1,459.975) = 279.448 kJ/mol

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