Question

Acetylene, C2H2, can be converted to ethane, C2H6, by a process known as hydrogenation. The reaction is C2H2(g)+2H2(g)?C2H6(g)

Given the following data, what is the value of Kp for this reaction?
Substance ΔG∘f
(kJ/mol)
C2H2(g) 209.2
H2(g) 0
C2H6(g) -32.89

Answer 1

Answer : The value of
`K_p` for this reaction is,
`2.73* 10^(42)`

Explanation :

The given chemical reaction is:

`C_2H_2(g)+2H_2(g)\rightarrow C_2H_6(g)`

Now we have to calculate value of
`(\Delta G^o)`.

`\Delta G^o=G_f_(product)-G_f_(reactant)`

`\Delta G^o=[n_(C_2H_6(g))* \Delta G^0_((C_2H_6(g)))]-[n_(C_2H_2(g))* \Delta G^0_((C_2H_2(g)))+n_(H_2(g))* \Delta G^0_((H_2(g)))]`

where,

`\Delta G^o` = Gibbs free energy of reaction = ?

n = number of moles

`\Delta G^0_((C_2H_6(g)))` = -32.89 kJ/mol

`\Delta G^0_((C_2H_2(l)))` = 209.2 kJ/mol

Now put all the given values in this expression, we get:

`\Delta G^o=[1mole* (-32.89kJ/mol)]-[1mole* (209.2kJ/mol)+2mole* (0kJ/mol)]`

`\Delta G^o=-242.09kJ/mol`

The relation between the equilibrium constant and standard Gibbs, free energy is:

`\Delta G^o=-RT* \ln K_p`

where,

`\Delta G^o` = standard Gibbs, free energy

R = gas constant = 8.314 J/L.atm

T = temperature = 298 K

`K_p` = equilibrium constant = ?

Now put all the given values in this expression, we get:

`-242.09kJ/mol=-(8.314J/L.atm)* (298K)* \ln K_p`

`K_p=2.73* 10^(42)`

Thus, the value of
`K_p` for this reaction is,
`2.73* 10^(42)`