Ethylene can be converted to ethanol according to the following reaction: C2H4(g

Question

Ethylene can be converted to ethanol according to the following reaction:

C2H4(g) + H2O(g) CH3CH2OH(l)

The reaction is catalyzed by H+(aq).

You will use the following table for this problem:

Substance:    Gf (kJ/mol)

C2H4(g)    72.7

H2O(g)     -225.1

CH3CH2OH(l) -175.7

Part 1:

At 298 Kelvin, what is Go (in kJ/mol) for this reaction?

Part 2:

Given your answer in part 1, which must be true about the value of K?

A. cannot be determined

B. K > 1

C. K < 1

Part 3:

What is the value of K for this reaction given your value of Go in part 1?

Explanation / Answer

part 1)

Balanced chemical equation is:

C2H4(g) + H2O(g) ---> CH3CH2OH(l)

Go rxn = 1*Gof(CH3CH2OH(l)) - 1*Gof( C2H4(g)) - 1*Gof(H2O(g))

Go rxn = 1*(-175.7) - 1*(72.7) - 1*(-225.1)

Go rxn = -23.3 KJ/mol

Answer: -23.3 KJ/mol

part 2)

since Go rxn is negative , K must be greater than 1

Answer: B. K>1

part 3)

Go rxn = -23.3 KJ/mol = -23300 J/mol

T = 298 K

use:

G = -R*T*ln Kc

-23300 = - 8.314*298* ln(Kc)

ln Kc = 9.404

Kc = 1.214*10^4

Answer: 1.214*10^4

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