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प्रश्न
Define the Enthalpy of ionisation.
उत्तर
Enthalpy of ionization is the enthalpy change accompanying the removal of an electron from one mole of a gaseous atom.
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संबंधित प्रश्न
Answer in brief.
How will you calculate reaction enthalpy from data on bond enthalpies?
Answer in brief.
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4CO(g) 2NO2(g) → 4CO2(g) + N2(g), ΔrH° = - 1200 kJ
Answer the following question.
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| ∆H°/kJ mol−1 | 414 | 243 | 431 |
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Calculate the standard enthalpy of formation of liquid methanol from the following data:
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Calculate the standard enthalpy of the reaction.
\[\ce{2Fe_{(s)} + \frac{3}{2} O_{2(g)} -> Fe2O_{3(s)}}\]
Given:
| 1. | \[\ce{2Al_{(s)} + Fe2O_{3(s)} -> 2Fe_{(s)} + Al_2O_{3(s)}}\], | ∆rH° = –847.6 kJ |
| 2. | \[\ce{2Al_{(s)} + \frac{3}{2} O_{2(g)} -> Al2O_{3(s)}}\], | ∆rH° = –1670 kJ |
When 2 moles of C2H6(g) are completely burnt, 3129 kJ of heat is liberated. If ∆Hf for CO2(g) and H2O(l) are −395 and −286 kJ per mole respectively, the heat combustion of C2H6(g) is ____________.
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Then, the heat of formation of CO is:
\[\ce{S + 3/2O2 -> SO3 +2{x} kcal}\] .........(i)
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\[\ce{C_{(s)} + O_{2(g)} -> CO_{2(g)}}\] ΔH° = -X kJ
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C = C bond energy: 606.10 kJ mol−1
C – C bond energy: 336.49 kJ mol−1
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Enthalpy for the given reaction will be:
\[\begin{array}{cc}
\phantom{}\ce{H}\phantom{...}\ce{H}\phantom{...................}\ce{H}\phantom{...}\ce{H}\phantom{....}\\
\phantom{.}|\phantom{....}|\phantom{....................}|\phantom{....}|\phantom{.....}\\
\ce{C = C + H - H -> H - C - C - H}\\
\phantom{.}|\phantom{....}|\phantom{....................}|\phantom{....}|\phantom{.....}\\
\phantom{}\ce{H}\phantom{...}\ce{H}\phantom{...................}\ce{H}\phantom{...}\ce{H}\phantom{....}
\end{array}\]
\[\ce{A -> B}\], ∆H = −10 kJ mol−1, Ea(f) = 50 kJ mol−1, then Ea of \[\ce{B -> A}\] will be ______.
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