(Solved): 1. a) Calculate the value of standard molar enthalpy and entropy of the chemical reactionat \( 298 ...

1. a) Calculate the value of standard molar enthalpy and entropy of the chemical reactionat \( 298 \mathrm{~K} \) using the data given: Is the chemical reaction exothermic reaction or not? WHY? \[ \mathrm{ZrO}_{2}(\mathrm{a})+2 \mathrm{CO}(\mathrm{g})(\mathrm{R}) \mathrm{Zr}(\mathrm{a})+2 \mathrm{CO}_{2}(\mathrm{~g}) \text { at } 298 \mathrm{~K} \] b) Calculate the enthalpy and entropy of chemical reaction at \( 1600 \mathrm{~K} \). Organize them. Do not calculate the values! \[ \mathrm{ZrO}_{2}(\beta)+2 \mathrm{CO}(\mathrm{g})(\mathrm{B}) \mathrm{Zr}(\beta)+2 \mathrm{CO}_{2}(\mathrm{~g}) \text { at } 1600 \mathrm{~K} \] DATA: \( D_{\mathrm{H} \mathrm{Zr}}(\mathrm{aR} \beta)=3900 \) joules at \( 1136 \mathrm{~K} \), \( \mathrm{DH} \mathrm{ZrO}_{2}(\mathrm{a} \AA \beta)=5900 \) joules at \( 1478 \mathrm{~K} \) Standard molar enthalpy formation and entropy at 298K: \[ \Delta \mathrm{H}_{1} \mathrm{ZrO}_{2}(\mathrm{a})=-1,100800 \text { joules, } \mathrm{SZr}(\mathrm{a})=39 \mathrm{~J} / \mathrm{K}, \quad \mathrm{SZrO}_{2}(\mathrm{a})=50.4 \] \( \mathrm{J} / \mathrm{K}, \quad \Delta \mathrm{H}_{\mathrm{t}} \mathrm{CO}(\mathrm{g})=-110500 \mathrm{~J}, \mathrm{SCO}(\mathrm{g})=197.5 \) \( \mathrm{J} / \mathrm{K}, \Delta \mathrm{HfCO}_{2}(\mathrm{~g})=-393500 \mathrm{~J}, \mathrm{SCO}_{2}(\mathrm{~g})=213.7 \mathrm{~J} / \mathrm{K} \)