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Crank
3 years ago
12

Calculate the entropy change for the reaction: Fe2O3(s) +3C(s) -> 2Fe(s) + 3CO(g)

Chemistry
1 answer:
Gelneren [198K]3 years ago
5 0

Answer:

ΔS = +541.3Jmol⁻¹K⁻¹

Explanation:

Given parameters:

Standard Entropy of Fe₂O₃ = 90Jmol⁻¹K⁻¹

Standard Entropy of C = 5.7Jmol⁻¹K⁻¹

Standard Entropy of Fe = 27.2Jmol⁻¹K⁻¹

Standard Entropy of CO = 198Jmol⁻¹K⁻¹

To find the entropy change of the reaction, we first write a balanced reaction equation:

                              Fe₂O₃ +  3C →  2Fe + 3CO

To calculate the entropy change of the reaction we simply use the equation below:

      ΔS = ∑S_{products} - ∑S_{reactants}

Therefore:

     ΔS = [(2x27.2) + (3x198)] - [(90) + (3x5.7)] = 648.4 - 107.1

                          ΔS = +541.3Jmol⁻¹K⁻¹

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Answer:

The answer to your question is  V2 = 1244 L

Explanation:

Data

Temperature 1 = 11°C

Volume = 1.03 x 10³ L

Temperature 2 = 70°C

Volume 2 = ?

To solve this problem use the Charles' law

        V1/ T1 = V2 / T2

-Solve for V2

        V2 = V1T2 / T1

Process

1.- Convert temperature to °K

Temperature 1 = 11°C + 273 = 284°K

Temperature 2 = 70°C + 273 = 343°K

2.- Substitution

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3.- Simplification

        V2 = 353290 / 284

4.- Result

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Answer:

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Explanation:

<u>Given the following data;</u>

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Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Charles' law is given by the formula;

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\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}

Where;

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\frac{V1}{T1} = \frac{V2}{T2}

Making V2 as the subject formula, we have;

V_{2}= \frac{V1}{T1} * T_{2}

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<em>hope it helps</em> ❤~

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