(Solved): 1. a) Calculate the value of standard molar enthalpy and entropy of the chemical reaction at \( 29 ...

1. a) Calculate the value of standard molar enthalpy and entropy of the chemical reaction at \( 298 \mathrm{~K} \) using the data given: Is the chemical reaction exothermic reaction or not? WHY? \[ \mathrm{ZrO}_{2}(\alpha)+2 \mathrm{CO}(\mathrm{g}) \rightarrow \underline{\mathrm{Zr}}(\alpha)+2 \mathrm{CO}_{2}(\mathrm{~g}) \text { at } 298 \mathrm{~K} \] b) Calculate the enthalpy and entropy of chemical reaction at \( 1600 \mathrm{~K} \). Organize them as we do it in the class. Do not calculate the values! \[ \mathrm{ZrO}_{2}(\beta)+2 \mathrm{CO}(\mathrm{g}) \rightarrow \mathrm{Zr}(\beta)+2 \mathrm{CO}_{2}(\mathrm{~g}) \text { at } 1600 \mathrm{~K} \] DATA: \( \Delta \mathrm{H} \mathrm{Zr}(\alpha \rightarrow \beta)=3900 \) joules at \( \mathbf{1 1 3 6 K} \), \( \Delta \mathrm{H} \mathrm{ZrO}_{2}(\alpha \rightarrow \beta)=5900 \) joules at \( \mathbf{1 4 7 8 K} \) Standard molar enthalpy formation and entropy at 298K: \[ \begin{array}{l} \Delta \mathrm{H}_{\mathrm{f}} \mathrm{ZrO}_{2}(\alpha)=-1,100800 \text { joules, } \mathrm{SZr}(\alpha)=39 \mathrm{~J} / \mathrm{K}, \quad \mathrm{SZrO}_{2}(\alpha)=50.4 \mathrm{~J} / \mathrm{K}, \quad \Delta \mathrm{H}_{\mathrm{r}} \mathrm{CO}(\mathrm{g})=-110500 \mathrm{~J}, \\ \mathrm{SCO}(\mathrm{g})=197.5 \mathrm{~J} / \mathrm{K}, \quad \Delta \mathrm{HfCO}_{2}(\mathrm{~g})=-393500 \mathrm{~J}, \mathrm{SCO}_{2}(\mathrm{~g})=213.7 \mathrm{~J} / \mathrm{K} \end{array} \]