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प्रश्न
Answer the following question.
State Hess’s law of constant heat summation. Illustrate with an example. State its applications.
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उत्तर
1. Hess’s law of constant heat summation:
Hess’s law of constant heat summation states that “Overall the enthalpy change for a reaction is equal to the sum of enthalpy changes of individual steps in the reaction”.
2. Illustration:
- The enthalpy change for a chemical reaction is the same regardless of the path by which the reaction occurs. Hess’s law is a direct consequence of the fact that enthalpy is a state function. The enthalpy change of a reaction depends only on the initial and final states and not on the path by which the reaction occurs.
- To determine the overall equation of the reaction, reactants and products in the individual steps are added or subtracted like algebraic entities.
- Consider the synthesis of NH3,
- \[\ce{2H_{2(g)} + N_{2(g)} → N2H_{4(g)}, Δ_r H^0_1 = + 95.4 kJ}\]
- \[\ce{N2H_{4(g)} + H_{2(g)} → 2NH_{3(g)}, Δ_r H^0_2 = -187.6 kJ}\]
\[\ce{H_{2(g)} + N_{2(g)} → 2NH_{3(g)}, Δ_rH^0 = - 92.2 kJ}\]
The sum of the enthalpy changes for steps (1) and (2) is equal to the enthalpy change for the overall reaction.
3. Application of Hess’s law:
The Hess's law has been useful to calculate the enthalpy changes for the reactions with their enthalpies being not known experimentally. To calculate heat of formation, combustion, neutralization, ionization, etc.
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Answer in brief.
How will you calculate reaction enthalpy from data on bond enthalpies?
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b) heat it to 100 °C and then
c) vapourise it at that temperature.
[Given: ΔfusH° (ice) = 6.01 kJ mol-1 at 0 °C, ΔvapH° (H2O) = 40.7 kJ mol-1 at 100 °C, Specific heat of water is 4.18 J g-1 K-1]
Define the Bond enthalpy.
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\[\ce{2Fe_{(s)} + \frac{3}{2} O_{2(g)} -> Fe2O_{3(s)}}\]
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 |
Write an application of Hess’s law.
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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lf, \[\ce{C_{(s)} + O2_{(g)} -> CO2_{(g)}}\], ∆H = x .........(i)
\[\ce{CO_{(g)} + 1/2O2_{(g)} -> CO2_{(g)}}\], ∆H = y .......(ii)
Then, the heat of formation of CO is:
Which among the following salts, solubility decreases with increase in temperature?
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Given that,
\[\ce{C_{(s)} + O_{2(g)} -> CO_{2(g)}}\] ΔH° = -X kJ
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From the following bond energies:
H – H bond energy: 431.37 kJ mol−1
C = C bond energy: 606.10 kJ mol−1
C – C bond energy: 336.49 kJ mol−1
C – H bond energy: 410.50 kJ mol−1
Enthalpy for the given reaction will be:
\[\begin{array}{cc}
\phantom{}\ce{H}\phantom{...}\ce{H}\phantom{...................}\ce{H}\phantom{...}\ce{H}\phantom{....}\\
\phantom{.}|\phantom{....}|\phantom{....................}|\phantom{....}|\phantom{.....}\\
\ce{C = C + H - H -> H - C - C - H}\\
\phantom{.}|\phantom{....}|\phantom{....................}|\phantom{....}|\phantom{.....}\\
\phantom{}\ce{H}\phantom{...}\ce{H}\phantom{...................}\ce{H}\phantom{...}\ce{H}\phantom{....}
\end{array}\]
Identify the invalid equation.
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\[\ce{H2O_{(l)}-> H2O_{(g)},}\] ΔVapH = 45.07 kJ mol−1.
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