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Answer the following question. Calculate ΔrH° for the following reaction at 298 K: 1) 2H3BO3(aq) → B2O3(s) + 3H2O(l), ΔrH° = + 14.4 kJ

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प्रश्न

Answer the following question.

Calculate ΔrH° for the following reaction at 298 K:

1) 2H3BO3(aq) → B2O3(s) + 3H2O(l), ΔrH° = + 14.4 kJ

2) H3BO3(aq) → HBO2(aq) + H2O(l), ΔrH° = - 0.02 kJ

3) H2B4O7(s) → 2B2O3(s) + H2O(l), ΔrH° = + 17.3 kJ

थोडक्यात उत्तर
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उत्तर

Given: Given equations are,

2H3BO3(aq) → B2O3(s) + 3H2O(l), ΔrH° = + 14.4 kJ   ...(i)

H3BO3(aq) → HBO2(aq) + H2O(l), ΔrH° = - 0.02 kJ    ....(ii)

H2B4O7(s) → 2B2O3(s) + H2O(l), ΔrH° = + 17.3 kJ   .....(iii)

To find: The standard enthalpy of the given reaction (ΔrH°)

Calculation:

Reverse equation (i) and multiply by 2,

2B2O3(s) + 6H2O(l) → 4H3BO3(aq) , ΔrH° = - 28.8 kJ ......(iv)

Multiply equation (ii) by 4

4H3BO3(aq) → 4HBO2(aq) + 4H2O(l), ΔrH° = - 0.08 kJ  .......(v)

Add equations (iv), (v) and (iii),

2B2O3(s) + 6H2O(l) → 4H3BO3(aq) , ΔrH° = - 28.8 kJ

4H3BO3(aq) → 4HBO2(aq) + 4H2O(l), ΔrH° = - 0.08 kJ

H2B4O7(s) → 2B2O3(s) + H2O(l), ΔrH° = + 17.3 kJ

_____________________________________________________

H2B4O7(s) + H2O(l) → 4HBO2(aq)  ΔrH° = - 28.8 + (- 0.08) + 17.3 = - 11.58 kJ

The standard enthalpy (ΔrH°) of the given reaction is -11.58 kJ.

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पाठ 4: Chemical Thermodynamics - Exercises [पृष्ठ ८९]

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बालभारती Chemistry [English] Standard 12 Maharashtra State Board
पाठ 4 Chemical Thermodynamics
Exercises | Q 4.17 | पृष्ठ ८९

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