If the H++OH-→ H2O+13.7kcal , then heat of complete neutralisation of 1 g mole H2SO4 with a base will be:

Question

If the H++OH-→ H2O+13.7kcal , then heat of complete neutralisation of 1 g mole H2SO4 with a base will be: (A)  13.7kcal  (B)  27.4kcal  (C)  6.85 text kcal  (D)  3.425kcal

Tardigrade Admin · Accepted Answer

1 mole of H2SO4 will be completely neutralized by two moles of OH- ion producing two moles of H2O . Hence, the amount of heat liberated will be 2× 13.7=27.4kcal underset1 mole mathopH2SO4 + underset2 moles mathop2OH- xrightarrowSO42-+ underset2 moles mathop2H2O +27.4kcal

Q. If the $H^{+} + O H^{-} \to H_{2} O + 13.7 k c a l$ , then heat of complete neutralisation of $1 g$ mole $H_{2} S O_{4}$ with a base will be:

$13.7 k c a l$

$27.4 k c a l$

$6.85 k c a l$

$3.425 k c a l$

1 mole of $H_{2} S O_{4}$ will be completely neutralized by two moles of $O H^{-}$ ion producing two moles of $H_{2} O$ . Hence, the amount of heat liberated will be $2 \times 13.7 = 27.4 k c a l$ $1 m o l e H_{2} S O_{4} + 2 m o l es 2 O H^{-} S O_{4}^{2 -} + 2 m o l es 2 H_{2} O + 27.4 k c a l$

Q. If the H++OH−→H2​O+13.7kcal , then heat of complete neutralisation of 1g mole H2​SO4​ with a base will be:

13.7kcal

27.4kcal

6.85kcal

3.425kcal