Q.

Given: 

 (1) HCO3(aq)+OH(aq)H2O(l)+CO32(aq):      ΔH1=41.84kJmol1 (2) H+(aq )+OH(aq)H2O(l):         ΔH2=57.32kJmol1

The enthalpy change for the reaction HCO3(aq)H+(aq)+CO32(aq) would be

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a

(41.8457.32) kJ mol1

b

(41.84+57.32) kJ mol1

c

(41.8457.32) kJ mol1

d

(41.84+57.32) kJ mol1

answer is B.

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Detailed Solution

Given equation is obtained by subtracting Eq. (2) from Eq. (1). Hence

ΔH=ΔH1ΔH2=(41.84+57.32) kJ mol1

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