Chemical Thermodynamics Question 254
Question: The combustion enthalpies of carbon, hydrogen and methane are $ -395.5,kJ,mo{{l}^{-1}} $ , $ -285.8,kJ,mo{{l}^{-1}} $ and $ -890.4,kJ,mo{{l}^{-1}} $ respectively at $ 25^{o}C $ . The value of standard formation enthalpies of methane at that temperature is [Pb. PMT 1998]
Options:
A) $ 890.4,kJ,mo{{l}^{-1}} $
B) $ -298.8,kJ,mo{{l}^{-1}} $
C) $ -74.7,kJ,mo{{l}^{-1}} $
D) $ -107.7,kJ,mo{{l}^{-1}} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ CH_4+2O_2\to CO_2+2H_2O,,\Delta H=-,890.4,kJ $ -.(i) $ C+O_2\to CO_2,\Delta H=-,395.5,kJ $ -..(ii) $ C+O_2\to CO_2,\Delta H=-,395.5,kJ $ .-.(ii) $ H_2+\frac{1}{2}O_2\to H_2O,\Delta H=-,285.8,kJ $ -..(iii) from (i), (ii), (iii). $ \Delta {H_{comb\text{.}}}(CH_4) $
$ =\Delta H_{f}(CO_2)+2\Delta H_{f}(H_2O)-\Delta H_{f}(CH_4)-2\Delta H_{f}(O_2) $
$ =-890.4=-395.5+2,(-285.5)-\Delta H_{f}(CH_4)-2\times 0 $
$ \Delta H_{f}(CH_4)=-74.7,kJ,mo{{l}^{-1}} $ .
