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rosijanka [135]
3 years ago
13

Iron(III) oxide can be reduced by carbon monoxide:Fe2O3 (s) + 3 CO (g) ⇌ 2 Fe (s) + 3 CO2 (g) Use the following thermodynamic da

ta at 298 K to determine the equilibrium constant at this temperature.SubstanceFe2O3 (s)CO (g)Fe (s)CO2 (g) ∆Hfº(kJ /mol)–824.2–110.50 –393.5∆Gfº(kJ/mol)–742.2 –137.20 –394.4Sº (J/mol•K)87.4 197.727.78213.7
Chemistry
1 answer:
Katena32 [7]3 years ago
3 0

Answer:

k = 1,423x10⁵

Explanation:

For the reaction:

Fe₂O₃(s) + 3CO (g) ⇌ 2Fe(s) + 3CO₂(g)

You can obtain equilibrium constant, k, using:

ΔG° = -RT lnK <em>(1)</em>

<em></em>

ΔG° of the reaction is:

ΔGf° products - ΔGf° reactants, that means:

ΔG° = 2∆Gfº Fe(s) + 3∆Gfº CO₂(g) - (∆Gfº Fe₂O₃(s) + 3∆Gfº CO (g)

ΔG° = 2×0 + 3×-394,4kJ/mol - (-742,2kJ/mol + 3×-137,2kJ/mol)

<em>ΔG° = -29,4 kJ/mol</em>

<em />

Replacing in (1) knowing R = 8,314472x10⁻³ kJ/molK; T = 298K

-29,4 kJ/mol = -8,314472x10⁻³ kJ/molK 298K lnK

11,866 = lnK

<em>1,423x10⁵ = k </em>

<em></em>

I hope it helps!

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Assoli18 [71]

Answer:

T2 =21.52°C

Explanation:

Given data:

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Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

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885J = 385 g× 1.1 J/g.°C×(T2 - 19.5°C )

885 J = 423.5 J/°C× (T2 - 19.5°C )

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T2 =21.52°C

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