H2SO4aq - 2KOHaq- K2SO4aq - 2H2Of - - H for the above reaction is-

The correct option is D 27.4 K.CalSolution: (C) 27.4 Kcal H_2SO_4_( aq. ) #160;+ 2 KOH_( aq. )⟶K_2SO_4_( aq. ) + 2 H_2O_( l ) According to re ...

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Standard XII

Chemistry

Heat of Reaction

H2SO4aq + 2KO...

Question

$H_{2} S O_{4 (a q)} + 2 K O H_{(a q)} \to K_{2} S O_{4 (a q)} + 2 H_{2} O_{(f)}; Δ H$ for the above reaction is:

- 13.7 K.Cal

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+57.3 K.J

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- 27.4 K.Cal

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- 137 K. J

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Solution

The correct option is D - 27.4 K.Cal
Solution:- (C) $- 27.4 K c a l$
${H_{2} S O_{4}}_{(a q .)} + 2 {K O H}_{(a q .)} ⟶ {K_{2} S O_{4}}_{(a q .)} + 2 {H_{2} O}_{(l)}$
According to reaction,
$1$ mole $H_{2} S O_{4}$ completely neutralised by $2$ mole of $K O H$ .
As we know that,
Gram equivalent $=$ no. of moles $\times$ Valency factor
Valency factor of $H_{2} S O_{4} = 2$
Therefore,
Gram equivalent of $H_{2} S O_{4} = 1 \times 2 = 2$
As we know that,
Heat of neutralisation of $1$ gm eq. of strong acid $= - 13.7 k J$
Heat of neutralisation of $2$ gm eq. of strong acid $= - 13.7 \times 2 = - 27.4 k c a l$
Hence $Δ H$ for the given reaction will be $- 27.4 k c a l$ .

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Similar questions

In the reaction the enthalpy change $K O H (a q) + H B r (a q) \to K B r (a q) + H_{2} O (l)$ ; $Δ H_{298 K} = - 13.7 K . c a l$ is referred as_______.

Δ H_{f} =

-98.2 K.Cal/mole

S_{N a} =

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I_{N a} =

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\frac{1}{2} D_{C l_{2}} =

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-184.2 K.Cal/mole
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[- E_{a}]

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Δ H^{0}

for the reaction

X_{(g)} + Y_{(g)} ⇌ Z (g)

- 4.6

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(R = 2 c a l . m o l^{- 1} K^{- 1})

The heat of combustion of $C H$ $_{4}$ (g), $C$ (graphite) and $H$ $_{4}$ (g) are respectively $- 20 k . c a l$ , $- 40 k . c a l$ and $- 10 k . c a l$ . The heat of formation of CH $_{4}$ is

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H2SO4(aq)+2KOH(aq)→K2SO4(aq)+2H2O(f);ΔH for the above reaction is:

$H_{2} S O_{4 (a q)} + 2 K O H_{(a q)} \to K_{2} S O_{4 (a q)} + 2 H_{2} O_{(f)}; Δ H$ for the above reaction is: