To calculate the Gibbs free energy change (ΔG) for the given reaction at a temperature of 290 K, we use the Gibbs free energy formula:
[tex]\[ \Delta G = \Delta H - T \Delta S \][/tex]
Given the values:
- [tex]\(\Delta H = -120 \text{ kJ}\)[/tex]
- [tex]\(\Delta S = -150 \text{ J/K}\)[/tex]
- Temperature ([tex]\(T\)[/tex]) = 290 K
First, we need to ensure that the units are consistent. To do this, we convert [tex]\(\Delta S\)[/tex] from J/K to kJ/K:
[tex]\[ \Delta S = -150 \, \text{J/K} = -150 \, \text{J/K} \times \frac{1 \, \text{kJ}}{1000 \, \text{J}} = -0.15 \, \text{kJ/K} \][/tex]
Now, using the Gibbs free energy formula, we substitute the given values:
[tex]\[ \Delta G = \Delta H - T \Delta S \][/tex]
[tex]\[ \Delta G = -120 \, \text{kJ} - 290 \, \text{K} \times (-0.15 \, \text{kJ/K}) \][/tex]
Next, perform the multiplication:
[tex]\[ 290 \, \text{K} \times (-0.15 \, \text{kJ/K}) = -43.5 \, \text{kJ} \][/tex]
Now, substitute this value back into our equation for [tex]\(\Delta G\)[/tex]:
[tex]\[ \Delta G = -120 \, \text{kJ} - (-43.5 \, \text{kJ}) \][/tex]
[tex]\[ \Delta G = -120 \, \text{kJ} + 43.5 \, \text{kJ} \][/tex]
[tex]\[ \Delta G = -76.5 \, \text{kJ} \][/tex]
Therefore, the calculated Gibbs free energy change (ΔG) at 290 K for the reaction [tex]\(2 NO_{(s)} + O_{2(g)} \rightarrow 2 NO_{2(g)}\)[/tex] is [tex]\(-76.5 \, \text{kJ}\)[/tex]. The entropy change [tex]\(\Delta S\)[/tex] converted to kJ/K is [tex]\(-0.15 \, \text{kJ/K}\)[/tex].
