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प्रश्न
The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, - 890.3 kJ mol-1 -393.5 kJ mol-1, and - 285.8 kJ mol-1 respectively. Enthalpy of formation of CH4(g) will be ______.
विकल्प
- 74.8 kJ mol-1
- 52.27 kJ mol-1
+ 74.8 kJ mol–1
+ 52.26 kJ mol-1
उत्तर
The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, - 890.3 kJ mol-1 -393.5 kJ mol-1, and - 285.8 kJ mol-1 respectively. Enthalpy of formation of CH4(g) will be - 74.8 kJ mol-1.
Explanation:
According to the question,
1) \[\ce{CH_{4(g)} + 2O_{2(g)} -> CO_{2(g)} + 2H_2O_{(g)}}\]
`triangle" H" = - 890.3 " kJ" " mol"^(-1)`
2) \[\ce{C_{(s)} + O_{2(g)} -> CO_{2(g)}}\]
`triangle" H" = - 393.5 " kJ mol"^(-1)`
3) \[\ce{2H_{2(g)} + O_{2(g)} -> 2H_2O_{(g)}}\]
`triangle" H" = - 285.8 " kJ mol"^(-1)`
Thus, the desired equation is the one that represents the formation of CH4(g) i.e.,
\[\ce{C_{(s)} + O_{2(g)} -> CO_{2(g)}}\]
`triangle_"f""H"_("CH"_4) = triangle_"c""H"_"c" + 2 triangle_"c" "H"_("H"_2) - triangle_"c""H"_("CO"_2)`
= [- 393.5 + 2(- 285.8) - (- 890.3)] kJ mol-1
= - 74.8 kJ mol-1
∴ Enthalpy of formation of CH4(g) = - 74.8 kJ mol-1
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Reason (R): The enthalpies of all elements in their standard state are zero.
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| `(Δ_"c""H"°)/"kJ mol"^-1` | −286 | −394 | −1560 |
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