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Write Brief Answer · Q51

Q.The partial pressure of carbon dioxide in the reaction
[!FORMULA] CaCO3(s)⇌CaO(s)+CO2(g)CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)
is 1.017×10−31.017 \times 10^{-3} atm at 5000500^0C. Calculate KPK_P at 6000600^0C for the reaction. ΔH\Delta H for the reaction is 181 kJ mol−1^{-1} and does not change in the given range of temperature.

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Step 1. For CaCO3(s)⇌CaO(s)+CO2(g)CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g), KP=pCO2K_P=p_{CO_2}, so KP1=1.017×10−3K_{P1}=1.017\times10^{-3} atm at T1=500+273=773T_1=500+273=773 K, and we want KP2K_{P2} at T2=600+273=873T_2=600+273=873 K, given ΔH=181×103\Delta H=181\times10^3 J mol−1^{-1}.

Step 2. Apply the integrated Van't Hoff equation: log⁡KP2KP1=ΔH2.303R(T2−T1T1T2)\log\dfrac{K_{P2}}{K_{P1}}=\dfrac{\Delta H}{2.303R}\left(\dfrac{T_2-T_1}{T_1T_2}\right).

Step 3. Compute the pieces: 2.303R=19.1472.303R=19.147; T2−T1=100T_2-T_1=100; T1T2=773×873=674,829T_1T_2=773\times873=674{,}829; so T2−T1T1T2=1.4819×10−4\dfrac{T_2-T_1}{T_1T_2}=1.4819\times10^{-4} K−1^{-1}. Then ΔH2.303R=18100019.147=9452.8\dfrac{\Delta H}{2.303R}=\dfrac{181000}{19.147}=9452.8. …

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