Question

delta S= 614

delta H = 3934.55

Discuss the values of delta H and delta S determined for Zn(s) I Zn^2+ (aq) II Pb^2+ (aq) I Pb(s) (analysis V. 1-2) in terms of

a. Whether the reaction is endothermic or exothermic

Answer 1

The exothermic or endothermic nature of a process is known from the sign of ΔH. In an exothermic reaction the heat energy is released. As a result the products have less enthalpy (heat content) than the reactants. Threfore

ΔH ( = Hp - Hr) is negative.

Thus for exothermic reactions ΔH = -ve

Similarly for Endothermic reactions ΔH = +ve

Since the value of ΔH for the reaction is 3934.55 (a positive value), it means the reaction is endothermic.

ALTERNATE EXPLANATION

For a reaction to proceed the ΔG must be negative.

ΔG = ΔH - TΔS

Since Δ is positive 614, the term TΔS is negative. Now if ΔH is greater than TΔS the reaction will not take place. In that case the reaction must be exothermic i.e, reaction has negative ΔH. However if TΔS is greater in magnitude than ΔH the reaction will proceed even if the reaction has positive value of ΔH. Such reactions are said to be entropy driven.