Question

Heat of neutralization of oxalic acid is –53.35 kJ mol^-1 using NaOH. Hence ΔH of

Answer 1

Answer:

It is given that Heat of neutralization of oxalic acid is –53.35 kJ mol-1 using NaOH. Hence ΔH of H2C2O4 ⇋ C2O2+4 + 2H+ is 7.9 kJ. Hence the correct answer is D. By the definition of the heat of neutralization we get,1/2 H2C2O4 + OH-à ½ C2O42-  + H2O , ΔH = -53.35 KJ –(i) H+  + OH- -> H2O. ΔH = -57.3 KJ. (ii). By (ii) – (i) we can get, ΔH = -3.95 * 2 = 7.9 KJ.