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Standard XI

Chemistry

Enthalpy

Enthalpy chan...

Question

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

Q. The enthalpy change for the following reaction,

N a O H (a q) + H C l (a q) ⟶ N a C l (a q) + H_{2} O (l)

is -57 kJ. Predict the value of the enthalpy change in the following reaction:

B a (O H)_{2} + H_{2} S O_{4} (a q) ⟶ B a S O_{4} (s) + 2 H_{2} O (l)

Q. Consider the following reactions.

(a)

H_{(a q)}^{+} + O H_{(a q)}^{-} = H_{2} O_{(l)}, Δ H = - X_{1}

m o l^{- 1}

(b)

H_{2 (g)} + \frac{1}{2} O_{2 (g)} = H_{2} O_{(l)}, Δ H = X_{2}

m o l^{- 1}

(c)

C O_{2 (g)} + H_{2 (g)} = C O_{(g)} + H_{2} O_{(l)} - X_{3}

m o l^{- 1}

(d)

C_{2} H_{2 (g)} + \frac{5}{2} O_{2 (g)} = 2 C O_{2 (g)} + H_{2} O_{(l)} + X_{4}

m o l^{- 1}

Enthalpy of formation of

H_{2} O_{(l)}

is:

Q. If enthalpy of neutralization of

{C H}_{3} C O O H

N a O H

- 49.86 k J / m o l

then enthalpy of ionization of

{C H}_{3} C O O H

is:

[Note : For strong acid and strong base neutralization net chemical change is :

H^{+} (a q) + {O H}^{-} (a q) ⟶ H_{2} O (l); Δ_{r} H^{o} = - 55.84 k J / m 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 $\times Δ H_{N}$

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