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Standard XII

Chemistry

Limiting Reagent or Reactant

500 ml of 1 M...

Question

500 ml of 1 M HCN is neutralised completely by 500ml of 1 M KOH solution and 1.36 kcal of heat is liberated. The $Δ$ H of dissociation of HCN is:

10.98 kcal/mol

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-10.98 kcal

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12.305 kcal/mol

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8.12 kcal/mol

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Solution

The correct option is B 10.98 kcal/mol
$n = \frac{500 \times 1}{1000} = 1 / 2$ mole = 0.5 mole

Heat released for 0.5 mol is 1.36kcal Then for $1 m o l e = 1.36 \times 2 = 2.72$ kcal

Heat change in dissociating HCN $= 13.7 k c a l - 2.72 k c a l = 10.98 k c a l$ .

Hence option A is correct.

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500 ml of 1 M HCN is neutralised completely by 500ml of 1 M KOH solution and 1.36 kcal of heat is liberated. The ΔH of dissociation of HCN is:

The correct option is B 10.98 kcal/moln=500×11000=1/2 mole = 0.5 moleHeat released for 0.5 mol is 1.36kcal Then for 1 mole=1.36×2=2.72 kcalHeat change in dissociating HCN =13.7 kcal−2.72 kcal=10.98 kcal.Hence option A is correct.

500 ml of 1 M HCN is neutralised completely by 500ml of 1 M KOH solution and 1.36 kcal of heat is liberated. The $Δ$ H of dissociation of HCN is:

The correct option is B 10.98 kcal/mol
$n = \frac{500 \times 1}{1000} = 1 / 2$ mole = 0.5 mole

Heat released for 0.5 mol is 1.36kcal Then for $1 m o l e = 1.36 \times 2 = 2.72$ kcal

Heat change in dissociating HCN $= 13.7 k c a l - 2.72 k c a l = 10.98 k c a l$ .

Hence option A is correct.