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Question
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
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Solution
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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