Question

given $\delta h^\circ = +257.7\\,\text{kj/mol}$ and $\delta s^\circ = +213.7\\,\text{j}\cdot\text{mol}^{-1}\text{k}^{-1}$, determine $\delta g^\circ$, in kj/mol, for the following reaction at $223\\,^\circ\text{c}$.

$$3\\,\\text{n}_2\\,(\text{g}) + 2\\,\\text{o}_2\\,(\text{g}) \ ightarrow 6\\,\\text{no}\\,(\text{g})$$

Explanation:

Step1: Convert temperature to Kelvin

To use the Gibbs free energy formula \(\Delta G^\circ=\Delta H^\circ - T\Delta S^\circ\), we first convert the temperature from Celsius to Kelvin. The formula for converting Celsius to Kelvin is \(T(K)=T(^\circ C)+ 273.15\). Given \(T = 223^\circ C\), so \(T=223 + 273.15=496.15\ K\).

Step2: Convert \(\Delta S^\circ\) units to kJ

The given \(\Delta S^\circ=+ 213.7\ J\cdot mol^{-1}K^{-1}\). To convert joules to kilojoules, we divide by 1000. So \(\Delta S^\circ=\frac{213.7}{1000}\ kJ\cdot mol^{-1}K^{-1}=0.2137\ kJ\cdot mol^{-1}K^{-1}\).

Step3: Calculate \(\Delta G^\circ\) using the formula

The formula for Gibbs free energy change is \(\Delta G^\circ=\Delta H^\circ - T\Delta S^\circ\). We know \(\Delta H^\circ = + 257.7\ kJ/mol\), \(T = 496.15\ K\) and \(\Delta S^\circ=0.2137\ kJ\cdot mol^{-1}K^{-1}\). Plugging in the values:
\[

$$\begin{align*} \Delta G^\circ&=257.7-(496.15\times0.2137)\\ &=257.7 - (496.15\times0.2137)\\ &=257.7-106.03\\ &=151.67\ kJ/mol \end{align*}$$

\]

Answer:

\(\boxed{151.7}\) (rounded to one decimal place)