Enthalpy change for the reaction $H_{(a q)}^{+} + O H_{(a q)}^{-} ⟶ H_{2} O_{(} l) I S - 57.3 k J m o l^{- 1} .$ If 1 L of $0.5 M J N O_{3}$ is mixed with 2L of 0.1 MKOH, the enthalpy change would be

-28.35 kJ

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28.65 kJ

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-11.46 kJ

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11.46 kJ

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Solution

The correct option is C
-11.46 kJ

Number of equiv. of $H N O_{3} = 1 \times 0.5 = 0.5;$ equivalents of $K O H = 2 \times 0.1 = 0.2$ (limiting reagent)

Enthalpy change = Number of equivalents $\times Δ H_{N}$

$= 0.2 \times (- 57.3) = 11.46 k J$

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Enthalpy change for the reaction $H_{(a q)}^{+} + O H_{(a q)}^{-} ⟶ H_{2} O_{(} l) I S - 57.3 k J m o l^{- 1} .$ If 1 L of $0.5 M J N O_{3}$ is mixed with 2L of 0.1 MKOH, the enthalpy change would be

H_{2 (g)} + \frac{1}{2} O_{2 (g)} \to H_{2} O_{(l)}

;

Δ H = - 286.2 k J

;

H_{2} O_{(l)} \to H_{(a q)}^{+} + O H_{(a q)}^{-}

;

Δ H = + 57.3 k J

. Enthalpy of ionisation of

O H^{-}

in aqueous solution is:

Q. Assertion :The enthalpy of formation of

H_{2} O (l)

is greater than that of

H_{2} O (g)

Reason: Enthalpy change is negative for the condensation reaction:

H_{2} O (g) ⟶ H_{2} O (l)

Q. Assertion :The enthalpy of formation of

H_{2} O (l)

is greater than that of

H_{2} O (g)

. Reason: Enthalpy change is negative for the condensation reaction

H_{2} O (g) ⟶ H_{2} O (l)

Assertion (A): The enthalpy of formation of $H_{2} O_{(l)}$ is greater than that of $H_{2} O (g)$

Reason (R): Enthalpy change is negative for the condensation reaction $H_{2} O (g) \to H_{2} O (l)$

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Enthalpy change for the reaction H+(aq)+OH−(aq)⟶H2O(l) IS−57.3 kJ mol−1. If 1 L of 0.5 MJNO3 is mixed with 2L of 0.1 MKOH, the enthalpy change would be

The correct option is C -11.46 kJ Number of equiv. of HNO3=1×0.5=0.5; equivalents of KOH=2×0.1=0.2 (limiting reagent) Enthalpy change = Number of equivalents ×ΔHN =0.2×(−57.3)=11.46 kJ

Enthalpy change for the reaction $H_{(a q)}^{+} + O H_{(a q)}^{-} ⟶ H_{2} O_{(} l) I S - 57.3 k J m o l^{- 1} .$ If 1 L of $0.5 M J N O_{3}$ is mixed with 2L of 0.1 MKOH, the enthalpy change would be

The correct option is C
-11.46 kJ

Number of equiv. of $H N O_{3} = 1 \times 0.5 = 0.5;$ equivalents of $K O H = 2 \times 0.1 = 0.2$ (limiting reagent)

Enthalpy change = Number of equivalents $\times Δ H_{N}$

$= 0.2 \times (- 57.3) = 11.46 k J$