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Fetre by calculaing \newlineΔ∘\Delta^\circ.\newlinecos⁡(s)+5(s)+2cos⁡(s)\cos(s)+5(s)+2\cos(s)\newlineDicrmine if the following reaction is spontaneous or not \newline25∘C25^\circ C by determining the free energg valhe.\newline\begin{align*}\(\newline&1. Decermine if the following reaction is spontancous of \ \newline&\quad \Delta S=-175.9\,\text{J}//\text{mol}\cdot\text{K},(\newline\)&2. \text{NO}(g)\rightarrow \text{N}_2\text{O}_4(g)\quad \Delta H=-57.2\,\text{kJ}//\text{mol}\quad\newline\end{align*}\)

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Q. Fetre by calculaing \newlineΔ∘\Delta^\circ.\newlinecos⁡(s)+5(s)+2cos⁡(s)\cos(s)+5(s)+2\cos(s)\newlineDicrmine if the following reaction is spontaneous or not \newline25∘C25^\circ C by determining the free energg valhe.\newline\begin{align*}\(\newline&1. Decermine if the following reaction is spontancous of \ \newline&\quad \Delta S=-175.9\,\text{J}//\text{mol}\cdot\text{K},(\newline\)&2. \text{NO}(g)\rightarrow \text{N}_2\text{O}_4(g)\quad \Delta H=-57.2\,\text{kJ}//\text{mol}\quad\newline\end{align*}\)
  1. Convert to Kelvin: Calculate the change in Gibbs free energy (ΔG\Delta G) using the formula ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S, where TT is the temperature in Kelvin. Convert 25∘C25^\circ C to Kelvin: 25+273.15=298.15 K25 + 273.15 = 298.15 \, K.
  2. Substitute values into formula: Substitute the values into the formula: ΔG=(−57.2 kJ/mol)−(298.15 K)(−175.9 J/mol⋅K)\Delta G = (-57.2 \, \text{kJ/mol}) - (298.15 \, \text{K})(-175.9 \, \text{J/mol}\cdot\text{K}). Convert J to kJ by dividing by 10001000: −175.9 J/mol⋅K=−0.1759 kJ/mol⋅K-175.9 \, \text{J/mol}\cdot\text{K} = -0.1759 \, \text{kJ/mol}\cdot\text{K}.
  3. Perform calculation: Perform the calculation: ΔG=−57.2 kJ/mol−(298.15 K)(−0.1759 kJ/mol⋅K)=−57.2 kJ/mol+52.45 kJ/mol.\Delta G = -57.2 \, \text{kJ/mol} - (298.15 \, \text{K})(-0.1759 \, \text{kJ/mol}\cdot\text{K}) = -57.2 \, \text{kJ/mol} + 52.45 \, \text{kJ/mol}.
  4. Calculate final value: Calculate the final value of ΔG\Delta G: −57.2 kJ/mol+52.45 kJ/mol=−4.75 kJ/mol-57.2 \, \text{kJ/mol} + 52.45 \, \text{kJ/mol} = -4.75 \, \text{kJ/mol}.

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