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Standard XII

Chemistry

Heat of Formation

The enthalpy ...

Question

The enthalpy of neutralization of 1M HCI by 1M NaOH is -57 kJ/mole. The enthalpy of formation of water is -285 kJ/mole. The enthalpy of formation of $O H^{-}$ ion is :

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Solution

$H C l + N a O H ⟶ N a C l + H_{2} O Δ H = - 57$ kJ
$Δ H_{n e u t r a l i s a t i o n} = {(Δ H)}_{H_{2} O} - {(Δ H_{f})}_{O H^{-}} ⟹ - 57 k J = - 285 k J - {(Δ H_{f})}_{O H^{-}}$
$⟹ Δ H_{f o r m a t i o n}$ of $O H^{-}$ $= - (285 - 57) = - 228 k J / m o l$

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The enthalpy of neutralization of 1M HCI by 1M NaOH is -57 kJ/mole. The enthalpy of formation of water is -285 kJ/mole. The enthalpy of formation of OH− ion is :

HCl+NaOH⟶NaCl+H2OΔH=−57 kJΔHneutralisation=(ΔH)H2O−(ΔHf)OH−⟹−57kJ=−285kJ−(ΔHf)OH−⟹ΔHformation of OH−=−(285−57)=−228kJ/mol

The enthalpy of neutralization of 1M HCI by 1M NaOH is -57 kJ/mole. The enthalpy of formation of water is -285 kJ/mole. The enthalpy of formation of $O H^{-}$ ion is :

$H C l + N a O H ⟶ N a C l + H_{2} O Δ H = - 57$ kJ
$Δ H_{n e u t r a l i s a t i o n} = {(Δ H)}_{H_{2} O} - {(Δ H_{f})}_{O H^{-}} ⟹ - 57 k J = - 285 k J - {(Δ H_{f})}_{O H^{-}}$
$⟹ Δ H_{f o r m a t i o n}$ of $O H^{-}$ $= - (285 - 57) = - 228 k J / m o l$