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Calculate the enthalpy change accompanying the transforming of C (graphite) to C(diamond) . Given that the enthalpies of combustion of graphite and diamond are 393.5 and 395.4 kJ `mol^(-1)` respectively. |
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Answer» We are given (i) C (graphite) `+ O_(2)(g) rarr CO_(2)(g) , Delta _(c ) H^(@)= - 393.5 kJ mol^(-1)` (ii) C (diamond) `+ O_(2)(g) rarr CO_(2)(g), Delta _(c ) H^(@) = - 395.4 kJ mol^(-1)` We aim at C( graphite ) `rarr ` C ( diamond) `, Delta _("trans") H^(@) = ?` Subtracting eqn. (ii) from eqn. (i), we get or C(graphite)- C (diamond) `rarr 0 , Delta _(r) H^(@) = - 393.5 - ( - 395.4) = + 1.9 kJ` |
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