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The reaction $A \leftrightharpoons B$; $\Delta H = +24\; kJ/mol$. For the reaction $B\leftrightharpoons C$; $\Delta H=-18$ kJ/mol. The decreasing order of enthalpy of $A, B$ and $C$ follow the order

(a) A, B, C
(b) B, C, A
(c) C, B, A
(d) C, A, B

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Answer: B, C, A
Let $\Delta H_f (A) = x$ kJ/mol
$A\longrightarrow B$ ; $\Delta H$ = + 24 kJ/mol
$\Delta H = \Delta _f H(B) - \Delta_fH(A)$
$\Rightarrow \Delta _f H(B) =\Delta H + \Delta_fH(A)$
$\Rightarrow \Delta_f H(B) = (x+24) $ kJ/mol
$B\longrightarrow C$ ; $\Delta H$ = - 18 kJ/mol
$\Delta H = \Delta _f H(C) - \Delta_fH(B)$
$\Rightarrow \Delta _f H(C) =\Delta H + \Delta_fH(B)$
$\Rightarrow \Delta_f H(B) = [-18+(x+24)] $ kJ/mol = $(x+6)$ kJ/mol
$\therefore$ In decreasing order $B>C>A$
answered Feb 27, 2014 by mosymeow_1

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