Question

Calculate the standard Gibbs free energy of reaction at 650 K for the reaction 1 CO(g) + \frac{1}{2} O_2(g) \to CO_2(g) by assuming that the standard molar enthalpy of the reaction is independent of the temperature. $\Delta_f H^\circ (CO, g) = -110 \text{kJ mol}^{-1}$ $\Delta_f H^\circ (CO_2, g) = -393.5 \text{kJ mol}^{-1}$ $\bar{S}^\circ (CO, g) = 197.7 \text{J K}^{-1} \text{mol}^{-1}$ $\bar{S}^\circ (O_2, g) = 205.2 \text{J K}^{-1} \text{mol}^{-1}$ $\bar{S}^\circ (CO_2, g) = 213.8 \text{J K}^{-1} \text{mol}^{-1}$ -257.2 kJ mol$^{-1}$ 25.77 kJ mol$^{-1}$ -226.8 kJ mol$^{-1}$ 334.3 kJ mol$^{-1}$ -283.0 kJ mol$^{-1}$

          Calculate the standard Gibbs free energy of reaction at 650 K for the reaction
1
CO(g) + \frac{1}{2} O_2(g) \to CO_2(g)
by assuming that the standard molar enthalpy of the reaction is independent of the
temperature.
$\Delta_f H^\circ (CO, g) = -110 \text{kJ mol}^{-1}$
$\Delta_f H^\circ (CO_2, g) = -393.5 \text{kJ mol}^{-1}$
$\bar{S}^\circ (CO, g) = 197.7 \text{J K}^{-1} \text{mol}^{-1}$
$\bar{S}^\circ (O_2, g) = 205.2 \text{J K}^{-1} \text{mol}^{-1}$
$\bar{S}^\circ (CO_2, g) = 213.8 \text{J K}^{-1} \text{mol}^{-1}$
-257.2 kJ mol$^{-1}$
25.77 kJ mol$^{-1}$
-226.8 kJ mol$^{-1}$
334.3 kJ mol$^{-1}$
-283.0 kJ mol$^{-1}$

Calculate the standard Gibbs free energy of reaction at 650 K for the reaction
1
CO(g) + (1)/(2) O2(g) →CO2(g)
by assuming that the standard molar enthalpy of the reaction is independent of the
temperature.
H^∘ (CO, g) = -110 kJ mol^-1
H^∘ (CO2, g) = -393.5 kJ mol^-1
S̅^∘ (CO, g) = 197.7 J K^-1mol^-1
S̅^∘ (O2, g) = 205.2 J K^-1mol^-1
S̅^∘ (CO2, g) = 213.8 J K^-1mol^-1
-257.2 kJ mol^-1
25.77 kJ mol^-1
-226.8 kJ mol^-1
334.3 kJ mol^-1
-283.0 kJ mol^-1

Added by Bradley F.

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