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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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3 0
3 years ago
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What pressure will be produced when 2.0 moles of N2 gas is heated to 68oC in a container that holds 1.25 of gas?
Stella [2.4K]

The pressure of the nitrogen gas produced is determined as 44.77 atm.

<h3>What is the pressure of the Nitrogen gas?</h3>

The pressure of the nitrogen gas is determined from ideal gas equation, as shown below;

PV = nRT

P = nRT/V

where;

  • n is number of moles = 2 moles
  • R is ideal gas constant = 0.08205 L.atm/mol.K
  • T is temperature = 68⁰C = 68 + 273 = 341 K
  • V is volume = 1.25 L

P = (2 x 0.08205 x 341)/(1.25)

P = 44.77 atm.

Learn more about pressure here: brainly.com/question/25736513

#SPJ1

3 0
2 years ago
Identify the group of elements that does not readily combine with other elements.
7nadin3 [17]
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also known as Inert gases
4 0
2 years ago
A sample containing 2.30 mol of Ne gas has an initial volume of 8.00 L. What is the final volume, in liters, when the following
marta [7]

Answer:

a. 4,00L

b. 16,00L

c. 12,31L

Explanation:

Avogadro's law says:

\frac{V_1}{n_1} =\frac{V_2}{n_2}

a. If initial conditions are 2,30mol and 8,00L and you lose one-half of atoms, that means you have 1,15mol:

\frac{8,00L}{2,30mol} =\frac{V_2}{1,15mol}

<em>V₂ = 4,00L</em>

b. If initial conditions are 2,30mol and 8,00L and you add 2,30mol, that means you have 4,60mol:

\frac{8,00L}{2,30mol} =\frac{V_2}{4,60mol}

<em>V₂ = 16,00L</em>

c. 25,0g of Ne are:

25,0g × (1mol / 20,1797g) = 1,24 moles of Ne. That means you have 2,30mol - 1,24mol = 3,54mol of Ne

\frac{8,00L}{2,30mol} =\frac{V_2}{3,54mol}

<em>V₂ = 12,31L</em>

I hope it helps!

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