Cr 2 O 3 (s)+3CO(g) --> 2Cr(s)+3CO 2 (g) The above reaction reaches equillibrium

Question

Cr2O3(s)+3CO(g) --> 2Cr(s)+3CO2(g)

The above reaction reaches equillibrium in a 2.0L flash. Data is collected at 25 Degrees C.

Cr2O3(s)+3CO(g) --> 2Cr(s)+3CO2(g)

Delta Heat (of formation) kJ/mol -1139.7 -110.5 0 -393.5

S (entropy) J/molK 81.2 197.7 3.8 213.8

Delta G( Gibbs energy) kJ/mol -1058.1 -137.2 0 -394.4

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1.) Reaction is exo or endo thermic?

2.) Is there an increase or decrease in entropy?

3.) What is the delta G at 25 degrees C? Is this spontaneous at 25 degrees C?

4.)What is the delta G at 2000 degrees C? Is the reaction spontaneous or nonspontaneous at 2000 degrees C?

5.) What effect does an increase in temperature have on the spontaneity of this reaction?

6.) At what temperature C does this reaction become spontaneous?

7.) What is the value of the equillibrium constant at 2000C?

8.) You initially only have Cr2O3 along with a partial pressure of CO equal to .5 atm in the 2L flask. What would the equillibrium partial pressure of CO2 be at 2000 degrees C? Use the equillibrium constant from question 7.

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Explanation / Answer

Cr2O3(s)+3CO(g) --> 2Cr(s)+3CO2(g)

We need

Hf units kJ/mol

Hf of Cr2O3(s) = ?1128.4

Hf of CO(g) = ?110.53

H of Cr(s) = 0 by definition

H of CO2(g) = ?393.51

Hf = Hproducts - Hreactants

H products = 2*Hf of Cr + 3*Hf of CO2 = 2*0 + 3*?393.51 = -1180.53 kJ

H reactants = 1*Hf of Cr2O3 + 3Hf of CO = 1*?1128.4 + 3* ?110.53 = -1459.99 kJ

Hf = Hproducts - Hreactants = -1180.53-(-1459.99) = 279.46 kJ

Hf = 279.46 kJ

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