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barxatty [35]
2 years ago
7

For a given reaction, ΔH=78.7kJ/mol, and the reaction is spontaneous at temperatures above the crossover temperature, 461K. The

value of ΔS= _______ J/mol⋅K, assuming that ΔH and ΔS do not vary with temperature.
Chemistry
1 answer:
TiliK225 [7]2 years ago
7 0

Answer:

171 J/mol.K

Explanation:

The spontaneity of a reaction depends on the Gibbs free energy (ΔG). If ΔG < 0, the reaction is spontaneous. The Gibbs free energy is related to the enthalpy (ΔH) and the entropy (ΔS) of the reaction through the following expression:

ΔG = ΔH - T.ΔS

where,

T is the absolute temperature

If ΔG < 0,

ΔH - T.ΔS < 0

ΔH < T.ΔS

ΔS > ΔH/T = (78.7 × 10³ J/mol)/461 K = 171 J/mol.K

ΔS must be at least 171 J/mol.K for the reaction to be spontaneous above 461 K.

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2c4H10 +13O2 —&gt;8CO2+10H2O to find how many moles of oxygen would react with 4.3 mol C4H10
Elena-2011 [213]

Answer:

\large\boxed{\text{28 mol of O$_{2}$}}

Explanation:

             2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

n/mol:      4.3  

13 mol of O₂ react with 2 mol of 2C₄H₁₀

\text{Moles of O}_{2} = \text{4.3 mol C$_{4}$H$_{10}$} \times \dfrac{\text{13 mol O}_{2}}{\text{2 mol C$_{4}$H$_{10}$}} = \textbf{28 mol O}_{2}\\\\\text{The reaction requires $\large\boxed{\textbf{28 mol of O$_{2}$}}$}

5 0
3 years ago
An electric range burner weighing 699.0 grams is turned off after reaching a temperature of 482.0°C, and is allowed to cool down
jasenka [17]

Answer:

0.42 J/gºC

Explanation:

We'll begin by calculating the heat energy used to heat up the water. This can be obtained as follow:

Mass (M) of water = 560 g

Initial temperature (T₁) = 22.7 °C

Final temperature (T₂) = 80.3 °C.

Specific heat capacity (C) of water = 4.18 J/gºC

Heat (Q) absorbed =?

Q = MC(T₂ – T₁)

Q = 560 × 4.18 (80.3 – 22.7)

Q = 2340.8 × 57.6

Q = 134830.08 J

Finally, we shall determine the specific heat capacity of the burner. This can be obtained as follow:

Mass (M) of burner = 699 g

Initial temperature (T₁) = 482.0°C

Final temperature (T₂) = 22.7 °C

Heat (Q) evolved = – 134830.08 J

Specific heat capacity (C) of the burner =?

Q = MC(T₂ – T₁)

–134830.08 = 699 × C (22.7 – 482.0)

–134830.08 = 699 × C × –459.3

–134830.08 = –321050.7 × C

Divide both side by –321050.7

C = –134830.08 / –321050.7

C = 0.42 J/gºC

Therefore, the specific heat capacity of the burner is 0.42 J/gºC

8 0
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Answer:

The percent composition for nitrogen in chromium(III) nitrate is 6%.

Explanation:

6 0
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Answer:

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Explanation:

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