The enthalpy change of the reaction is <u>-1347.8 kJ.</u>
<h3>What is the enthalpy change, ΔH, of the reaction?</h3>
The enthalpy change, ΔH, of the reaction is calculated from Hess's law of constant heat summation as follows:
Hess's law states that the enthalpy change of a reaction is the sum of the enthalpies of the intermediate reaction.
Given the reactions below and their enthalpy values;
1. X (s) + 12 O₂ (g)⟶ XO (s) ΔH₁ = −850.5 kJ
2. XCO₃ (s) ⟶ XO (s) + CO₂ (g) ΔH₂ = +497.3 kJ
The enthalpy change, ΔH, of the reaction whose equation is given below, will be:
X (s) + 12 O₂ (g) + CO₂ (g) ⟶ XCO₃ (s)
ΔH = ΔH₁ - ΔH₂
ΔH = − 850.5 kJ - (+497.3 kJ)
ΔH = -1347.8 kJ
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Answer:
X is 3-chloro-2,4-dimethylpentane
Explanation:
The radical chlorination of 2,4-dimethylpentane may give three products.
Based on the information that
i) both the products gives single alkene on elimination and
ii) both the products also undergoes SN2 reaction but Y reacts faster than X
We may conclude that as SN2 reaction is faster in primary alkyl halides as compared to secondary or tertiary so Y must be primary or secondary and it cannot be tertiary alkyl halide.
The possible products are shown in the figure.
The structure of X is shown in the figure.
603.040 has 5 significant figures in it I'm pretty sure
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