Consider the reaction:

Chemistry

1 answer:

Vesna [10]2 years ago

7 0

Answer:

The correct answer is E.

Explanation:

Given the reaction:

C₂H₅OH (l) + 3 O₂(g) ⇒ 2 CO₂ (g) + 3 H₂O (l) ; ΔH = -1.37×10³ kJ

We can see that ΔH is negative. This means that the product formation energy is less than the reagent formation energy. In other words, the reaction releases heat and is therefore an exothermic reaction.

In addition, the enthalpy of formation of liquid water has different value than the enthalpy of formation of gaseous water. Therefore, the enthalpy variation would have a different value if the water formed was in a gaseous state.

Finally, the propositions II and III are true.

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Answer:

Explanation:

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3 years ago

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Keith_Richards [23]

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2 years ago

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The molar heat capacity of ethane is represented in the temperature range 298 K to 400 K by the empirical expression Cp,m in J K

BabaBlast [244]

Answer:

-88.66 kJ/mol

Explanation:

The expressions of heat capacity (Cp,m) for C(s) and for H₂(g) are:

C(s): Cp,m/(J K-1 mol-1) = 16.86 + (4.77T/10³) - (8.54x10⁵/T²)

H₂(g): Cp,m/(J K-1 mol-1) = 27.28 + (3.26T/10³) + (0.50x10⁵/T²)

Cp = A + BT + CT⁻²

For the Kirchoff's Law:

ΔHf = ΔH°f + $\int\limits^{T2}_{T1} {DCp(T)} \, dT$

Where ΔH°f is the enthalpy at 298 K, T1 is 298 K, T2 is the temperature given (373 K), and DCp is the variation of Cp (products less reactants). ΔH°f for ethene is -84.68 kJ/mol and the reaction is:

2C(s) + 3H₂(g) → C₂H₆

So, DCp:

dA = A(C₂H₆) - [2xA(C) + 3xA(H₂)] = 14.73 - [2x16.86 + 3x27.28] = -100.83

dB = B(C₂H₆) - [2xB(C) + 3xB(H₂)] = 0.1272 - [2x4.77x10⁻³ + 3x3.26x10⁻³] = 0.10788

dC = C(C₂H₆) - [2xC(C) + 3xC(H₂)] = 0 - (2x(-8.54x10⁵) + 3x0.50x10⁵) = 15.58x10⁵

dCp = -100.83 + 0.10788T + 15.58x10⁵T⁻²

$\int\limits^{373}_{298} {-100.83 + 0.10788T + 15.58x10^5T^{-2}} \, dT$ = -3796.48 J/mol = -3.80 kJ/mol (solved by a graphic calculator)