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Standard XII

Chemistry

Hess' Law

A: Enthalpy o...

Question

A: Enthalpy of neutralization of 1 equivalent each of HCI and $H_{2} S O_{4}$ with NaOH is same.
R: Enthalpy of neutralization is always the heat evolved when 1 mole acid is neutralized by a base

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Solution

$A \to$ 1 equivalent has same enthalpy of nuetralization as both mean enthalpy of $H^{+} + O H^{-} ⇌ H_{2} O$

$R \to$ It is wrong becuase it happened due to the equivalent system. $Δ_{n e t} H$ does not mean only are $H^{+}$

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The enthalpy of neutralization of $H C l$ and $N a O H$ is $- 57 k J m o l^{- 1}$ . The heat evolved at constant pressure (in kJ) when $0.5 m o l e$ of H2SO4 react with $0.75 m o l e$ of $N a O H$ is equal to

Q. The heat of neutralization is the change in enthalpy that occurs when one equivalent of an acid and one equivalent of a base undergo a neutralization reaction to form water and a salt. Heat of neutralization of strong acid with strong base is

- 13.7 k c a l m o l^{- 1}

\begin{matrix} List - I & List - II (|Enthalpy change in kcal|) & (Neutralisation) (I) < 13.7 kcal & (P) H C l 1 mol + N a O H 1 mol (II) = 13.7 kcal & (Q) H F 1 mol + N a O H 1 mol (III) > 13.7 kcal & (R) H C l 1 mol + N H_{4} O H 1 mol (IV) = 27.4 kcal & (S) N a O H 2 mol + H_{2} S O_{4} 1 mol (T) N a O H 1 mol + C H_{3} C O O H 1 mol \end{matrix}

Which of the following options has the correct combination considering List-I and List-II?

- 13.7 k c a l m o l^{- 1}

\begin{matrix} List - I & List - II (|Enthalpy change in kcal|) & (Neutralisation) (I) < 13.7 kcal & (P) H C l 1 mol + N a O H 1 mol (II) = 13.7 kcal & (Q) H F 1 mol + N a O H 1 mol (III) > 13.7 kcal & (R) H C l 1 mol + N H_{4} O H 1 mol (IV) = 27.4 kcal & (S) N a O H 2 mol + H_{2} S O_{4} 1 mol (T) N a O H 1 mol + C H_{3} C O O H 1 mol \end{matrix}

Which of the following options has the correct combination considering List-I and List-II?

Q. The enthalpy of neutralization of oxalic acid by a strong base is

- 25.4 k

c a l {m o l}^{- 1}

. The enthalpy of neutralization of strong acid and strong base is

- 13.7 k c a l

{e q}^{- 1}

. The enthalpy of dissociation of oxalic acid is:
[Given: heat of neutralization will be

(2 \times - 13.7 k c a l)

]

Q. The enthalpy of neutralization of one mole of

H_{3} P O_{3}

acid using two moles of

N a O H

- 106.68 {kJ.mol}^{- 1}

. If the enthalpy of neutralization of

H C l

N a O H

- 55.84 {kJ.mol}^{- 1}

, calculate the

Δ H_{ionization}

H_{3} P O_{3}

into its ions:

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A: Enthalpy of neutralization of 1 equivalent each of HCI and H2SO4 with NaOH is same.R: Enthalpy of neutralization is always the heat evolved when 1 mole acid is neutralized by a base

A→ 1 equivalent has same enthalpy of nuetralization as both mean enthalpy of H++OH−⇌H2OR→ It is wrong becuase it happened due to the equivalent system. ΔnetH does not mean only are H+

A: Enthalpy of neutralization of 1 equivalent each of HCI and $H_{2} S O_{4}$ with NaOH is same.
R: Enthalpy of neutralization is always the heat evolved when 1 mole acid is neutralized by a base

$A \to$ 1 equivalent has same enthalpy of nuetralization as both mean enthalpy of $H^{+} + O H^{-} ⇌ H_{2} O$

$R \to$ It is wrong becuase it happened due to the equivalent system. $Δ_{n e t} H$ does not mean only are $H^{+}$