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  • delta U of combustion of methane is minus X kJ mol raised to minus 1. The value of delta H is

6.4   \Delta U ^\ominusof combustion of methane is - X KJ mol ^{-1}. The value of \Delta H ^\ominus is
             (i) = \Delta U ^\ominus
             (ii)> \Delta U ^\ominus
             (iii) < \Delta U ^\ominus
             (iv) = 0

Answers (1)

Equation of combustion of methane-

CH_4(g)+\:2O_2(g)\rightarrow CO_2(g)\:+\:2H_2O(l)

Since, \Delta H^o=\Delta U^o+\Delta n_gRT and \Delta U^o=-X\:KJ\:mol^{-1}

from the equation \Delta n_g=(n_p-n_R)=1-3=-2

\Delta H^o=-X-2RT

\Delta H^o<\Delta U^o

so the correct option is (iii).

 

Posted by

Ankit

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