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Question
Calculate ∆H° for the following reaction:
2H3BO3(aq) → B2O3(s) + 3H2O(l)
a) H3BO3(aq) → HBO2(aq) + H2O(l) , ∆`H_1^@` = − 0.02 kJ
b) H2B4O7(s) → 2B2O3(s) + H2O(l) , ∆`H_2^@` = 17.3 kJ
c)H2B4O7(s) + H2O(l) → 4HBO2(aq), ∆`H_3^@` = − 11.58 kJ
Solution
Given: Given equations are
H3BO3(aq) → HBO2(aq) + H2O(l) , ∆`H_1^@` = − 0.02 kJ ....(i)
H2B4O7(s) → 2B2O3(s) + H2O(l) , ∆`H_2^@` = 17.3 kJ....(ii)
H2B4O7(s) + H2O(l) → 4HBO2(aq), ∆`H_3^@` = − 11.58 kJ.....(iii)
To find: The standard enthalpy of the reaction, ΔH°
Calculation: Multiply equation (i) by (2),
2H3BO3(aq) ⎯→ 2HBO2(aq) + 2H2O(l), ΔH° = −0.04 kJ …. (iv)
Multiply equation (ii) by ½
∴The ΔH° of the reaction = +14.4 kJ
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