Question

consider the reaction, hq ⇌ h⁺(aq) + q⁻(aq) at 298 k, the δs_rxn = -81.0 j/mol·k and the δg_rxn of 76.0 kj/mol. calculate δh_rxn in kj/mol.

Explanation:

Step1: Recall the Gibbs - Helmholtz equation

The Gibbs - Helmholtz equation is given by \(\Delta G=\Delta H - T\Delta S\), where \(\Delta G\) is the Gibbs free energy change, \(\Delta H\) is the enthalpy change, \(T\) is the temperature in Kelvin, and \(\Delta S\) is the entropy change. We need to solve for \(\Delta H\), so we can rearrange the formula to \(\Delta H=\Delta G+T\Delta S\).

Step2: Convert the units of \(\Delta S\)

We know that \(\Delta S = - 81.0\space J/mol\cdot K\) and we need to convert it to \(kJ/mol\cdot K\) because \(\Delta G\) is in \(kJ/mol\). Since \(1\space kJ = 1000\space J\), we have \(\Delta S=\frac{- 81.0\space J/mol\cdot K}{1000\space J/kJ}=- 0.081\space kJ/mol\cdot K\). The temperature \(T = 298\space K\) and \(\Delta G=76.0\space kJ/mol\).

Step3: Substitute the values into the formula for \(\Delta H\)

Substitute \(\Delta G = 76.0\space kJ/mol\), \(T = 298\space K\) and \(\Delta S=-0.081\space kJ/mol\cdot K\) into the formula \(\Delta H=\Delta G + T\Delta S\).

\[

$$\begin{align*} \Delta H&=76.0\space kJ/mol+298\space K\times(- 0.081\space kJ/mol\cdot K)\\ &=76.0\space kJ/mol-298\times0.081\space kJ/mol\\ &=76.0\space kJ/mol - 24.138\space kJ/mol\\ &=51.862\space kJ/mol\approx51.9\space kJ/mol \end{align*}$$

\]

Answer:

The value of \(\Delta H_{rxn}\) is approximately \(51.9\space kJ/mol\) (or \(51.862\space kJ/mol\) depending on the level of precision required).