At 25 ∘c , the equilibrium partial pressures for the reaction were found to be pa=5.16 bar, pb=5.04 bar, pc=4.11 bar, and pd=4.85 bar . a(g)+2b(g)↽−−⇀4c(g)+d(g) what is the standard change in gibbs free energy of this reaction at 25 ∘c ? δ∘rxn= kjmol

5.85kJ/Kmol.

Explanation:

The balanced equilibrium reaction is

The expression for equilibrium reaction will be,

Now put all the given values in this expression, we get the concentration of methane.

Relation of standard change in Gibbs free energy and equilibrium constant is given by:

where,

R = universal gas constant = 8.314 J/K/mole

T = temperature =

= equilibrium constant = 10.6

Thus standard change in Gibbs free energy of this reaction is 5.85kJ/Kmol.

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml