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Consider the following reaction : CO_((g)) + 2H_(2(g)) hArr CH_(3)OH_((g)) Given : Delta_(r) H^(@) (CH_(3)OOH, g) = -201 kJ/mol, Delta_(r) H^(@) (CO, g) = -114 kJ/mol S^(@) (CH_(3)OOH, g) = 240 J/K-mol , S^(@) (H_(2), g) = 29 JK^(-1)mol^(-1) S^(@) (CO, g) = 198 J/mol-K , C^(@)_(p,m) (H_(2)) = 28.8 J/mol-K C^(@)_(p,m) (CO) = 29.4 J/mol-K , C^(@)_(p,m) (CH_(3)OH) = 44 J/mol-K and ln ((320)/(300)) = 0.06, all data at 300 K Delta_(r) H^(@) at 300 K for the reaction is : |
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Answer» `-87` kJ/mol `Delta_(r ) H^(0) = Delta_(F) H^(0) (CH_(3)OH) - Delta_(f)H^(0) (CO) - 2. Delta_(f)H^(0) (H_(2))` `= -87kJ//mol` `Delta_(r) S-(320)^(0) - Delta_(r) S_(300)^(0) = Delta_(r)C_(P) [T_(2) - T_(1)]` where `Delta_(r) C_(P)^(0) = 44-29.4 -2 xx 28.8 = -43J//K - mol` `Delta_(r) S_(320)^(0) = - 16 + (-43) "ln "(320)/(320) = - 18.58` `Delta_(r)H_(320)^(0) = Delta_(r)H_(300)^(0) + Delta_(r) C_(P)^(0) [T_(2) - T_(1)]` `= - 87 + ((-43) xx 20)/(1000) = - 87.86` kJ/mol `Delta_(r) G_(320)^(0) = Delta_(r) H_(320)^(0) - T. Delta_(r) S_(320)^(0)` `= -87.86 - (320 xx (-18.58))/(1000) = - 81.91 kJ//mol` |
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