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6.14  Calculate the standard enthalpy of formation of CH3OH(l) from the following data:

    CH_3OH (l) + 3/2 O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \Delta _r H = -726 kJ mol^{-1}\\\\ .\: \: \: \: C(graphite) + O_2(g) \rightarrow CO_2(g) ;\Delta_ cH = -393 kJ mol ^ {-1}\\\\ .\: \: \: H_2(g) + 1/2 O_2(g) \rightarrow H_2O(l) ; \Delta_ f H = - 286 kJ mol^{-1}.

Answers (1)

best_answer

for the formation of CH_3OH the reaction is ,

C+2H_{2}O+\frac{1}{2}O_{2}\rightarrow CH_3OH
This can be obtained by the following expressions-
required eq = eq (i) + 2 (eq iii) - eq (i)

 \Delta _fH[CH_{3}OH] = \Delta _cH^\Theta +2\Delta _fH[H_{2}O]-\Delta _rH^\Theta
                                    =(-393) + 2(-286)-(-726)
                                     =-239 kJmol^{-1}

Posted by

manish

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