Question

using the data from the previous steps, calculate the standard gibbs free energy for the no₂ reaction at 25⁰c. 2no₂(g) → 2no(g) + o₂(g) δh°ᵣₓₙ = +112 kj δs°ᵣₓₙ = +147 j/k δg°ᵣₓₙ = ? kj enter either a + or - sign and the magnitude.

Explanation:

Step1: Convert temperature to Kelvin

The temperature is \(25^\circ\text{C}\). To convert to Kelvin, we use \(T = 25 + 273.15 = 298.15\,\text{K}\) (we can use \(298\,\text{K}\) for simplicity).

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

Given \(\Delta S^\circ_{\text{rxn}} = + 147\,\text{J/K}\). To convert to kJ/K, we divide by 1000: \(\Delta S^\circ_{\text{rxn}}=\frac{147}{1000}=0.147\,\text{kJ/K}\)

Step3: Use the Gibbs Free Energy formula

The formula for standard Gibbs Free Energy is \(\Delta G^\circ_{\text{rxn}}=\Delta H^\circ_{\text{rxn}} - T\Delta S^\circ_{\text{rxn}}\)
Substitute the values: \(\Delta H^\circ_{\text{rxn}} = 112\,\text{kJ}\), \(T = 298\,\text{K}\), \(\Delta S^\circ_{\text{rxn}} = 0.147\,\text{kJ/K}\)
\(\Delta G^\circ_{\text{rxn}}=112-(298\times0.147)\)
First, calculate \(298\times0.147 = 298\times0.1 + 298\times0.04 + 298\times0.007=29.8+11.92 + 2.086 = 43.806\)
Then, \(\Delta G^\circ_{\text{rxn}}=112 - 43.806 = 68.194\approx68\,\text{kJ}\) (or more precisely, using \(T = 298.15\))
\(T\Delta S^\circ=298.15\times0.147 = 298.15\times0.1+298.15\times0.04 + 298.15\times0.007=29.815+11.926+2.08705 = 43.82805\)
\(\Delta G^\circ=112 - 43.82805 = 68.17195\approx68\,\text{kJ}\) (or if we keep more decimals, but the answer is approximately \(+ 68\) kJ)

Answer:

\(+68\) (or more precisely, around \(+68.2\) but \(+68\) is acceptable based on significant figures or calculation steps)