For a gaseous reaction, standard conditions are 298 k and a partial pressure of 1 bar for all species. for the reaction
2 no ( g ) + o 2 ( g ) > 2 no 2 ( g )
the standard change in gibbs free energy is δ g ° = − 32.8 kj / mol . what is δ g for this reaction at 298 k when the partial pressures are:
pno = 0.500 bar , po2 = 0.250 bar , and pno 2 = 0.800 bar
deltag = ?

Answer : The value of is, -27.0kJ/mole

Explanation :

The formula used for is:

  ............(1)

where,

= Gibbs free energy for the reaction

=  standard Gibbs free energy  = -32.8 kJ

/mol

R = gas constant = 8.314 J/mole.K

T = temperature = 298 K

= equilibrium constant

First we have to calculate the value of .

The given balanced chemical reaction is,

The expression for equilibrium constant will be :

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

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

Now we have to calculate the value of by using relation (1).

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

Therefore, the value of is, -27.0kJ/mole

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