The standard free energy change for the formation of two moles of H20 (l) in a s

Question

The standard free energy change for the formation of two moles of H20 (l) in a strong acid-strong base neutralization reaction at 25 degrees C is -79.9 kJ.

H30+(aq)+OH-(aq)--->2H20(l) delta G=-79.9 kJ

a. Calculate the equilibrium constant for the reaction.

b. The standard enthalpy change delta H, for a strong acid-strong base reaction is -57.8k J. Determine the standard entropy change delta S for the reaction at 25 degrees C.

c. Explain the chemical significance of the calculated entropy change for the neutralization reaction

Explanation / Answer

(a) G=-RTlnK, here G=-79.9kJ/mol, R=0.008314kJ*mol-1K-1. T=298K. So plug those in:

-79.9=-(0.008314)(298)lnK. So K=1.013e14.

(b) G=H-TS. Here you have G=-79.9kJ/mol, H=-57.8kJ/mol, T=298K. So plug those in:

-79.9=-57.8-298*S; so S=0.0742kJ/(mol*K).

(c) the entropy change is positive, so the entropy increases; this makes a chemical reaction more spontaneous.

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