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प्रश्न
Does the following reaction represent a thermochemical equation?
\[\ce{CH_{4(g)} + 2O_{2(g)} -> CO_{2(g)} + 2H2O_{(g)}}\], ∆fH° = –900 kJ mol–1
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उत्तर
No, the given reaction does not represent a thermochemical equation. The given standard enthalpy of reaction (∆rH°) indicates the enthalpy change accompanying the reaction when the reactants and products involved are in their standard states. The standard state of water is H2O(l) and not H2O(g).
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संबंधित प्रश्न
Answer in brief.
How will you calculate reaction enthalpy from data on bond enthalpies?
Answer the following question.
Calculate ΔrH° for the following reaction at 298 K:
1) 2H3BO3(aq) → B2O3(s) + 3H2O(l), ΔrH° = + 14.4 kJ
2) H3BO3(aq) → HBO2(aq) + H2O(l), ΔrH° = - 0.02 kJ
3) H2B4O7(s) → 2B2O3(s) + H2O(l), ΔrH° = + 17.3 kJ
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| Bond | C − H | Cl − Cl | H − Cl |
| ∆H°/kJ mol−1 | 414 | 243 | 431 |
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Which of the following quantities of heat energy will be evolved when 39 g C6H6 are burnt?
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H – H bond energy: 431.37 kJ mol−1
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C – H bond energy: 410.50 kJ mol−1
Enthalpy for the given reaction will be:
\[\begin{array}{cc}
\phantom{}\ce{H}\phantom{...}\ce{H}\phantom{...................}\ce{H}\phantom{...}\ce{H}\phantom{....}\\
\phantom{.}|\phantom{....}|\phantom{....................}|\phantom{....}|\phantom{.....}\\
\ce{C = C + H - H -> H - C - C - H}\\
\phantom{.}|\phantom{....}|\phantom{....................}|\phantom{....}|\phantom{.....}\\
\phantom{}\ce{H}\phantom{...}\ce{H}\phantom{...................}\ce{H}\phantom{...}\ce{H}\phantom{....}
\end{array}\]
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Calculate ΔsubH of the H2O from the given data:
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\[\ce{H2O_{(l)}-> H2O_{(g)},}\] ΔVapH = 45.07 kJ mol−1.
Calculate the standard enthalpy of combustion of methane if the standard enthalpy of formation of methane, carbon dioxide and water are −74.8, −393.5 and −285.8 kJmol−1 respectively.
