InterviewSolution
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InterviewSolution
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It is planned to carry the reaction `CaCO_(3)(s) hArr CaO (s) +CO_(2)(g)` at 1273K and 1 bar (a) Is this reaction spontaneous at this temperature and pressure ? (b) Calculate the value of (i) `K_(p)` at 1273 K for the reaction and (ii) partial pressure of `CO_(2)` at equilibrium Given `Delta_(f) H^(@)` values (`kJ mol^(-1)) :` `CaCO_(3)(s) =-1206.9, CaO(s)= - 635.1, CO_(2)(g) = - 393.5` `S^(@) ` value `(JK^(-1) mol^(-1))` `CaCO_(3)(s) =92.9, CaO(s) = 39.8, CO_(2)(g) = 213.7`. |
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Answer» (a) `Delta_(r) H^(@) = Delta_(f) H^(@) ( CaO) +Delta_(f)H^(@) ( CO_(2))]-[Delta_(f)H^(@) ( CaCO_(3))]` `= [ - 635.1 + ( -393.5) ] - ( - 1206.9 )= + 178.3 kJ mol^(-1)` `Delta_(r)S^(@) = [S^(@) ( CaO) + S^(@) ( CO_(2)) ] - [S^(@)(CaCO_(3))]` `= ( 39.8 +213 .7)- 92.9= + 160.6 JK^(-1) mol^(-1)` `Delta_(r)G^(@) =Delta_(r)H^(@) - T Delta_(r)S^(@)` `= 178300 Jmol^(-1) - 1273 K xx 160.6 JK^(-1) mol^(-1)` `= - 26143.8 J mol^(-1) = 26144 mol^(-1) = 26.1 kJ mol^(-1)` As `Delta_(r) G^(@)` is negative, hence the reaction is spontaneous (b)(i) `DeltaG^(@) = - 2.303 RT log K_(p)` or `log K_(p) = ( - 26144 J mol^(-1))/( 2.303 xx 8.314 JK^(-1) xx 1273 K )` `= 1.0726` `K_(p) =` Antilog `1.0726 = 11.82` Applying law of chemical equilibrium `K_(p) = p_(CO_(2)) = 11.82` bar |
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