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Softa [21]
1 year ago
13

For the reaction of sodium bromide with chlorine gas

Chemistry
1 answer:
lina2011 [118]1 year ago
7 0

The reaction of sodium bromide with chlorine gas is Cl₂(aq) + 2Na + 2Br? 2Na + 2Cl⁻ + Br²(aq).

<h3>What is sodium bromide?</h3>

Sodium bromide is an inorganic compound, white, crystalline with high melting point.

The reaction between halogens is redox reaction

Oxidation  – 2Br⁻ ? Br₂ + 2e⁻ loss of electron.

Reduction – Cl₂ + 2e⁻ ? 2Cl⁻ gains of electron.

Thus, the correct option is Cl₂(aq) + 2Na + 2Br? 2Na + 2Cl⁻ + Br²(aq).

Learn more about sodium bromide

brainly.com/question/15409724

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Look at the following enthalpy diagram. Select all that apply.
Drupady [299]

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Option 2 and 4 are correct

Explanation:

The reactants in the attached image have more enthalpy and hence less stability as they are more reactive. Thus, Product is more stable than the reactants.

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Reactants have more energy than the products.  

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2 years ago
Calculate the enthalpy of the reaction below (∆Hrxn, in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g).
nalin [4]

The enthalpy change of the reaction below (ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.

The bond energies data is given as follows:

BE  for C≡O  = 1072 kJ/mol

BE for Cl-Cl = 242 kJ/mol

BE for C-Cl = 328 kJ/mol

BE for C=O = 766 kJ/mol

The enthalpy change for the reaction is given as :

ΔHr×n = ∑H reactant bond - ∑H product bond

ΔHr×n = ( BE C≡O + BE Cl-Cl) - ( BE C=O + BE 2 × Cl-Cl )

ΔHr×n = ( 1072 + 242 ) - ( 766 + 656 )

ΔHr×n = 1314 - 1422

ΔHr×n = - 108 kJ

Thus, The enthalpy change of the reaction below ( ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.

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