Question

A student determines the value of the equilibrium constant to be 1.43×1064 for the following reaction. 2Na(s) + 2H2O(l)2NaOH(aq) + H2(g) Based on this value of Keq: G° for this reaction is expected to be (greater, less) than zero. Calculate the free energy change for the reaction of 2.31 moles of Na(s) at standard conditions at 298K. G°rxn = kJ

Answer 1

solution:

1)

Since K > 1 then G° < 0

We know that G° = -RT InK

= - (8.314 J/K/mol) ( 298 K) In (1.43×10⁶⁴)

= -366060.6 J

  = - 366.06 kJ

the balanced reaction is:

2Na(s) + 2H₂O(l) -------------> 2NaOH(aq) + H₂(g)   G°rxn =  - 366.06 kJ

2 mol of Na in a reaction, G°rxn =  - 366.06 kJ

then for 2.31 mol of Na = (2.31 mol/ 2 mol) x  - 366.06 kJ

G°rxn = -422.79 kJ

Therefore,

G°rxn = - 422.79 kJ

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