What is the ΔG° at room temperature of the reaction 2CO+O2→2CO2, if ΔH°=-128.3 kJ and ΔS°= - 159.5 J/K ?*- 80. 769 kJ80. 769 kJ- 175.831 kJ- 124.313 kJ
Question
What is the ΔG° at room temperature of the reaction 2CO+O2→2CO2, if ΔH°=-128.3 kJ and ΔS°= - 159.5 J/K ?*- 80. 769 kJ80. 769 kJ- 175.831 kJ- 124.313 kJ
Solution
To calculate the ΔG° at room temperature for the reaction 2CO+O2→2CO2, we can use the Gibbs free energy equation:
ΔG° = ΔH° - TΔS°
Given that ΔH° = -128.3 kJ (or -128300 J to match the units of ΔS°) and ΔS° = -159.5 J/K, we can substitute these values into the equation.
Assuming room temperature to be 298.15 K (approximately 25°C), the equation becomes:
ΔG° = -128300 J - 298.15 K * -159.5 J/K
Solving this gives:
ΔG° = -128300 J + 47531.175 J
ΔG° = -80768.825 J
Converting this back to kJ gives:
ΔG° = -80.769 kJ
So, the ΔG° at room temperature for the reaction 2CO+O2→2CO2 is -80.769 kJ.
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