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Liono4ka [1.6K]
3 years ago
6

Trichloroethylene, a widely used degreasing solvent for machine parts, is produced in a two-step reaction sequence. Ethylene is

first chlorinated to yield tetrachloroethane, which is dehydrochlorinated to form trichloroethylene.
C2H4 (g) + 2Cl2 (g) → C2H2Cl4 (l) + H2 (g): AHᵣ⁰ = -385.76 kJ
C2H2Cl4 (l) → C2HCl3 (l) + HCl (g)

The standard heat of formation of liquid trichloroethylene is -276.2 kJ/mol.
Use the given data and tabulated standard heats of formation of ethylene and HCl to calculate the standard heat of formation of tetrachloroethane and the standard heat of reaction for the second reaction.
Chemistry
1 answer:
solmaris [256]3 years ago
4 0

Answer:

ΔH°C2H2Cl4(l) = -333,36 kJ/mol

ΔH°r₂ = -35,14 kJ/mol

Explanation:

The ΔH°r of the first reaction is:

ΔH°r = -385,76 kJ/mol = (ΔH°C2H2Cl4(l) + ΔH°H2(g)) - (ΔH°C2H4(g) + 2ΔHCl2 (g))

ΔH°H2(g) = 0 kJ/mol

ΔH°C2H4(g) = 52,4 kJ/mol

Δ°HCl2 (g) = 0 kJ/mol

Replacing:

ΔH°C2H2Cl4(l) = -385,76 kJ/mol + 52,4 kJ/mol = <em>-333,36 kJ/mol</em>

The standard heat of the second reaction is:

ΔH°r₂ = ΔH°C2HCl3(l) + ΔH°HCl(g) - ΔH°C2H2Cl4(l)

Where:

ΔH°C2HCl3(l) = -276,2 kJ/mol; ΔH°HCl(g) = -92,3 kJ/mol; ΔH°C2H2Cl4(l) = -333,36 kJ/mol

Replacing:

ΔH°r₂ = -276,2 kJ/mol -92,3 kJ/mol + 333,36 kJ/mol

<em>ΔH°r₂ = -35,14 kJ/mol</em>

<em></em>

I hope it helps!

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Calculate the molar concentration of OH ions in a 0.066 M solution of ethylamine (C2H5NH2: Kb= 6.4 x 10-)
wel

Answer:

6.5x10⁻³M = [OH⁻]

Explanation:

The Kb of a Weak base as ethylamine is expressed as follows:

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As the equilibrium of ethylenamine is:

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Also, we can assume the concentration of ethylamine doesn't decrease. Replacing:

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6.5x10⁻³M = X

<h3>6.5x10⁻³M = [OH⁻]</h3>
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