The enthalpy of dissociation of HCN computed from the following data, is
HCl(aq) + NaOH(aq) $⟶$ NaCl(aq) + H₂O; $Δ H_{1} = - 55.9 k J / m o l e$
HCN(aq) + NaOH(aq) $⟶$ NaCN(aq) + H₂O; $Δ H_{2} = - 12.1 k J / m o l e$

Question

The enthalpy of dissociation of HCN computed from the following data, is

HCl(aq) + NaOH(aq) $⟶$ NaCl(aq) + H₂O; $Δ H_{1} = - 55.9 k J / m o l e$

HCN(aq) + NaOH(aq) $⟶$ NaCN(aq) + H₂O; $Δ H_{2} = - 12.1 k J / m o l e$

TestHub · Accepted Answer

Δ H 2 ​ − Δ H 1 ​ = − 12.1 − ( − 55.9 ) = 43.8 k J .

Chemistry - Thermochemistry Question with Solution | TestHub