Answers
(a)
Thermochemical equation :
HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) ∆H= -2.134kJ
Heat released by neutralization of acid = m.s.∆T
= (33+42)ml×1g/ml × 4.184J/g.℃ × (31.8-25.0)℃ = 2133.84 J
= 2.13384kJ = 2.134kJ
∆H = -2.134kJ (Answer)
(b)
Heat released due to neutralization of 33ml , 1.20M HCl
Number of moles of HCl = 1.20mol/L × 33×10-3L
= 3.96×10-2mol
∆H = - 2.13384kJ/(3.96×10-2mol)
= -53.8848kJ/mol
= - 53.88 kJ/mol. (Answer)
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