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Question

# A 200 mL solution of NaOH requires 100 mL of 0.44 M aqueous HCl for complete neutralization. The temperature of the mixture increased by 1.96 K . Calculate the enthalpy change for the neutralization of one mole of hydrogen ions. Given specific heat of water =4.184 J/g K

A
56 kJ
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B
40 kJ
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C
90 kJ
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D
110 kJ
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Solution

## The correct option is A −56 kJSince specific heat of solution is not given assuming dilute solution have the same specific heat capacity as does water. q=mcsΔT=300×4.18×(1.96)=2.46 kJ Volume of solution=100+200mL=300 mL Weight of solution=300 g This heat flow is used to calculate ΔH for 1 mole of the reaction HCl(aq)+NaOH(aq)→NaCl(aq)+H2O(l) Moles of HCl reacted =0.1×0.44==0.044 mol 1 mol HCl yield 1 mole of H+ 0.044 mol HCl yield 0.044 mole of H+ H−+OH→H2O(l) Neutralization of 0.044 mole of H+ generates 2.46 kJ of heat. The reaction is exothermic ΔH=−ve4.4×10−2 mole H+(aq)=−2.46kJ1 mole H+(aq)=−2.464.4×10−2=−56kJΔHreaction=−56 kJ

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