Question

at what temperature, in °c, is a certain reaction at equilibrium if δh = +86.4 kj/mol and δs = +170.2 j·mol⁻¹·k⁻¹?

Explanation:

Step1: Recall the equilibrium condition for Gibbs free energy

At equilibrium, \(\Delta G = 0\). The formula relating \(\Delta G\), \(\Delta H\), \(\Delta S\), and temperature \(T\) is \(\Delta G=\Delta H - T\Delta S\). Setting \(\Delta G = 0\), we get \(0=\Delta H - T\Delta S\), which can be rearranged to \(T=\frac{\Delta H}{\Delta S}\).

Step2: Convert units of \(\Delta H\)

Given \(\Delta H = + 86.4\space kJ/mol\), we need to convert it to \(J/mol\) to match the unit of \(\Delta S\) (which is in \(J\cdot mol^{-1}\cdot K^{-1}\)). Since \(1\space kJ = 1000\space J\), \(\Delta H=86.4\times1000 = 86400\space J/mol\).

Step3: Calculate the temperature in Kelvin

Substitute \(\Delta H = 86400\space J/mol\) and \(\Delta S=170.2\space J\cdot mol^{-1}\cdot K^{-1}\) into the formula \(T = \frac{\Delta H}{\Delta S}\). So \(T=\frac{86400\space J/mol}{170.2\space J\cdot mol^{-1}\cdot K^{-1}}\approx507.6\space K\).

Step4: Convert temperature from Kelvin to Celsius

The formula to convert Kelvin to Celsius is \(T(^{\circ}C)=T(K)- 273.15\). Substituting \(T(K) = 507.6\space K\), we get \(T(^{\circ}C)=507.6 - 273.15\approx234.45^{\circ}C\).

Answer:

The temperature at which the reaction is at equilibrium is approximately \(\boldsymbol{234^{\circ}C}\) (or more precisely around \(234.5^{\circ}C\) depending on rounding during calculation).